Inter 1st Year

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b)

All problems in this exercise have reference to ΔABC.

I.

Question 1.
Express \(\Sigma r_{1} \cot \frac{A}{2}\) in terms of s.
Solution:
\(\Sigma r_{1} \cot \frac{A}{2}\) = \(\Sigma\left(s \tan \frac{A}{2}\right) \cot \frac{A}{2}\)
= Σs
= s + s + s
= 3s

Question 2.
Show that Σa cot A = 2(R + r).
Solution:
L.H.S = Σa . cot A
= Σ2R sin A \(\frac{\cos A}{\sin A}\)
= 2R Σ cos A
= \(2 R\left(1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)\) (From transformants)
= \(2\left(R+4 R \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right)\)
= 2(R + r)
= R.H.S

Question 3.
In ∆ABC, prove that r₁ + r₂ + r₃ – r = 4R.
Solution:

Question 4.
In ∆ABC, prove that r + r₁ + r₂ – r₃ = 4R cos C.
Solution:

Question 5.
If r + r₁ + r₂ + r₃ then show that C = 90°.
Solution:

II.

Question 1.
Prove that 4(r₁r₂ + r₂r₃ + r₃r₁) = (a + b + c)²
Solution:

Question 2.
Prove that \(\left(\frac{1}{r}-\frac{1}{r_{1}}\right)\left(\frac{1}{r}-\frac{1}{r_{2}}\right)\left(\frac{1}{r}-\frac{1}{r_{3}}\right)=\frac{a b c}{\Delta^{3}}=\frac{4 R}{r^{2} s^{2}}\)
Solution:

Question 3.
Prove that r(r₁ + r₂ + r₃) = ab + bc + ca – s².
Solution:

Question 4.
Show that \(\sum \frac{r_{1}}{(s-b)(s-c)}=\frac{3}{r}\)
Solution:

Question 5.
Show that \(\left(r_{1}+r_{2}\right) \tan \frac{C}{2}=\left(r_{3}-r\right) \cot \frac{C}{2}=c\)
Solution:

Question 6.
Show that r₁r₂r₃ = \(r^{3} \cot ^{2} \frac{A}{2} \cdot \cot ^{2} \frac{B}{2} \cdot \cot ^{2} \frac{C}{2}\)
Solution:

III.

Question 1.
Show that cos A + cos B + cos C = 1 + \(\frac{r}{R}\)
Solution:
L.H.S = cos A + cos B + cos C
= 2 cos(\(\frac{A+B}{2}\)) cos(\(\frac{A-B}{2}\)) + cos C

Question 2.
Show that \(\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2}=2+\frac{r}{2 R}\)
Solution:

Question 3.
Show that \(\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=1-\frac{r}{2 R}\)
Solution:

Question 4.
Show that
(i) a = (r₂ + r₃) \(\sqrt{\frac{r r_{1}}{r_{2} r_{3}}}\)
(ii) ∆ = r₁r₂ \(\sqrt{\frac{4 R-r_{1}-r_{2}}{r_{1}+r_{2}}}\)
Solution:

Question 5.
Prove that \(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r^{2}\) = 16R² – (a² + b² + c²).
Solution:

Question 6.
If p₁, p₂, p₃ are altitudes drawn from vertices A, B, C to the opposite sides of a triangle respectively, then show that
(i) \(\frac{1}{p_{1}}+\frac{1}{p_{2}}+\frac{1}{p_{3}}=\frac{1}{r}\)
(ii) \(\frac{1}{p_{1}}+\frac{1}{p_{2}}-\frac{1}{p_{3}}=\frac{1}{r_{3}}\)
(iii) p₁ . p₂ . p₃ = \(\frac{(a b c)^{2}}{8 R^{3}}=\frac{8 \Delta^{3}}{a b c}\)
Solution:

Question 7.
If a = 13, b = 14, c = 15, show that R = \(\frac{65}{8}\), r = 4, r₁ = \(\frac{21}{2}\), r₂ = 12 and r₃ = 14.
Solution:
a = 13, b = 14, c = 15
s = \(\frac{a+b+c}{2}\)
= \(\frac{13+14+15}{2}\)
= 21
s – a = 21 – 13 = 8
s – b = 21 – 14 = 7
s – c = 21 – 15 = 6

Question 8.
If r₁ = 2, r₂ = 3, r₃ = 6 and r = 1, prove that a = 3, b = 4 and c = 5.
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(a)

All problems in this exercise refer to ΔABC

I.

Question 1.
Show that Σa(sin B – sin C) = 0
Solution:

Question 2.
If a = √3 + 1 cms, ∠B = 30°, ∠C = 45°, then find c.
Solution:
∠B = 30°, ∠C = 45° and a = (√3 + 1) cms
A = 180° – (B + C)
= 180° – (30° + 45°)
= 180° – 75°
= 105°

Question 3.
If a = 2 cms, b = 3 cms, c = 4 cms, then find cos A.
Solution:
a = 2 cms, b = 3 cms and c = 4 cms

Question 4.
If a = 26 cms, b = 30 cms and cos C = \(\frac{63}{65}\), then find c.
Solution:
a = 26 cms, b = 30 cms and cos C = \(\frac{63}{65}\)
c² = a² + b² – 2ab cos C
⇒ c² = 676 + 900 – 2 × 26 × 30 × \(\frac{63}{65}\)
⇒ c² = 1576 – 1512
⇒ c² = 64
⇒ c = 8

Question 5.
If the angles are in the ratio 1 : 5 : 6, then find the ratio of its sides.
Solution:
Given \(\frac{A}{1}=\frac{B}{5}=\frac{C}{6}\), B = 5A, C = 6A
A + B + C = 180°
⇒ A + 5A + 6A = 180°
⇒ 12A = 180°
⇒ A = 15°
∴ B = 5A = 75°
∴ C = 6A = 90°
a : b : c = sin A : sin B : sin C
= sin 15° : sin 75° : sin 90°
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}: \frac{\sqrt{3}+1}{2 \sqrt{2}}: 1\)
= (√3 – 1) : (√3 + 1) : 2√2

Question 6.
Prove that 2(bc cos A + ca cos B + ab cos C) = a² + b² + c².
Solution:
L.H.S = Σ2bc cos A
= Σ2bc \(\frac{\left(b^{2}+c^{2}-a^{2}\right)}{2 b c}\)
= Σ(b² + c² – a²)
= b² + c² – a² + c² + a² – b² + a² + b² – c²
= a² + b² + c²
= R.H.S

Question 7.
Prove that \(\frac{a^{2}+b^{2}-c^{2}}{c^{2}+a^{2}-b^{2}}=\frac{\tan B}{\tan C}\)
Solution:
L.H.S = \(\frac{a^{2}+b^{2}-c^{2}}{c^{2}+a^{2}-b^{2}}=\frac{\tan B}{\tan C}\)
[∵ c² = a² + b² – 2ab cos C and b² = a² + c² – 2ac cos B]

Question 8.
Prove that (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c
Solution:
L.H.S = (b + c) cos A + (c + a) cos B + (a + b) cos C
= (b cos A + c cos A) + (c cos B + a cos B) + (a cos C + b cos C)
= (b cos C + c cos B) + (a cos C + c cos A) + (a cos B + b cos A)
= a + b + c
= R.H.S

Question 9.
Prove that (b – a cos C) sin A = a cos A sin C
Solution:
LHS = (b – a cos C) sin A
= (a cos C + c cos A – a cos C) sin A
= c cos A sin A [∵ b = a cos C + c cos A]
= (2R sin C) cos A sin A
= a cos A sin C (∵ 2R sin A = a)

Question 10.
If 4, 5 are two sides of a triangle and the included angle is 60°, find its area.
Solution:
Let a = 4, b = 5 are two sides and included angle is C = 60° then
Area of ∆ = \(\frac{1}{2}\) ab sin C
= \(\frac{1}{2}\) × 4 × 5 × sin 60°
= 2 × 5 × \(\frac{\sqrt{3}}{2}\)
= 5√3 sq.cm

Question 11.
Show that \(b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2}=s\)
Solution:

Question 12.
If \(\frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C}\), then show that ∆ABC is equilateral.
Solution:
Given that \(\frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C}\)
⇒ \(\frac{2 R \sin A}{\cos A}=\frac{2 R \sin B}{\cos B}=\frac{2 R \sin C}{\cos C}\)
⇒ \(\frac{\sin A}{\cos A}=\frac{\sin B}{\cos B}=\frac{\sin C}{\cos C}\)
⇒ tan A = tan B = tan C
⇒ A = B = C
⇒ ∆ABC is an equilateral triangle.

II.

Question 1.
Prove that a cos A + b cos B + c cos C = 4R sin A sin B sin C.
Solution:
L.H.S = (2R sin A) cos A + (2R sin B) cos B + (2R sin C) cos C
= R (sin 2A + sin 2B + sin 2C)
= R (2 sin (A + B) cos (A – B) + sin 2C)
= R [2 sin (180° – C) cos (A – B) + sin 2C]
= R (2 sin C . cos (A – B) + 2 sin C . cos C)
= 2R sin C (cos (A – B)) + cos C)
= 2R sin C (cos (A – B) + cos (180° – \(\overline{\mathrm{A}+\mathrm{B}}\))
= 2R sin C [cos (A – B) – cos (A + B)]
= 2R sin C (2 sin A sin B)
= 4R sin A sin B sin C
= R.H.S.

Question 2.
Prove that Σa³ sin(B – C) = 0.
Solution:
L.H.S = Σa² [a sin (B – C)]
= Σa² [2R . sin A sin (B – C)]
= R Σ a² (2 sin (180° – \(\overline{\mathrm{B}+\mathrm{C}}\)) sin (B – C))
= R Σ a² [2 sin (B + C) . sin (B – C)]
= R Σ a² (sin² B – sin² C)
= R Σ a² \(\left(\frac{b^{2}}{4 R^{2}}-\frac{c^{2}}{4 R^{2}}\right)\)
= \(\frac{1}{2 R}\) Σ[a² (b² – c²)]
= \(\frac{1}{2 R}\) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)]
= \(\frac{1}{2 R}\) (a²b² – a²c² + b²c² – a²b² + a²c² – b²c²)
= \(\frac{1}{2 R}\) × 0
= 0
= R.H.S

Question 3.
Prove that \(\frac{a \sin (B-C)}{b^{2}-c^{2}}=\frac{b \sin (C-A)}{c^{2}-a^{2}}=\frac{c \sin (A-B)}{a^{2}-b^{2}}\)
Solution:

Question 4.
Prove that \(\sum a^{2} \frac{a^{2} \sin (B-C)}{\sin B+\sin C}=0\)
Solution:

Question 5.
Prove that \(\frac{a}{b c}+\frac{\cos A}{a}=\frac{b}{c a}+\frac{\cos B}{b}=\frac{c}{a b}+\frac{\cos C}{c}\)
Solution:

Question 6.
Prove that \(\frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}=\frac{a^{2}+b^{2}}{a^{2}+c^{2}}\)
Solution:

Question 7.
If C = 60°, then show that
(i) \(\frac{\mathbf{a}}{\mathbf{b}+\mathbf{c}}+\frac{\mathbf{b}}{\mathbf{c}+\mathbf{a}}=1\)
(ii) \(\frac{b}{c^{2}-a^{2}}+\frac{a}{c^{2}-b^{2}}=0\)
Solution:
∠C = 60°
⇒ c² = a² + b² – 2ab cos C
⇒ c² = a² + b² – 2ab cos 60°
⇒ c² = a² + b² – 2ab (\(\frac{1}{2}\))
⇒ c² = a² + b² – ab

Question 8.
If a : b : c = 7 : 8 : 9, find cos A : cos B : cos C.
Solution:

Question 9.
Show that \(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}\)
Solution:

Question 10.
Prove that (b – a) cos C + c (cos B – cos A) = c . sin(\(\frac{A-B}{2}\)) cosec(\(\frac{A+B}{2}\))
Solution:
L.H.S = b cos C – a cos C + c cos B – c cos A
= (b cos C + c cos B) – (a cos C + c cos A)
= a – b
= 2R (sin A – sin B)

Question 11.
Express \(a \sin ^{2} \frac{C}{2}+c \cdot \sin ^{2} \frac{A}{2}\) interms of s, a, b, c.
Solution:

Question 12.
If b + c = 3a, then rind the value of \(\cot \frac{\mathrm{B}}{2} \cot \frac{\mathrm{C}}{2}\)
Solution:

Question 13.
Prove that (b + c) cos (\(\frac{B+C}{2}\)) = a cos (\(\frac{B-C}{2}\))
Solution:

Question 14.
In ΔABC, show that \(\frac{b^{2}-c^{2}}{a^{2}}=\sin \frac{(B-C)}{(B+C)}\)
Solution:

III.

Question 1.
Prove that
(i) \(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\frac{s^{2}}{\Delta}\)
(ii) \(\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}\) = \(\frac{b c+c a+a b-s^{2}}{\Delta}\)
(iii) \(\frac{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}{\cot A+\cot B+\cot C}\) = \(\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\)
Solution:

Question 2.
Show that
(i) Σ(a + b) tan(\(\frac{A-B}{2}\)) = 0
(ii) \(\frac{\mathbf{b}-\boldsymbol{c}}{\mathbf{b}+\mathbf{c}} \cot \frac{A}{2}+\frac{b+c}{b-c} \tan \frac{A}{2}\) = 2 cosec(B – C)
Solution:

Question 3.
(i) If sin θ = \(\frac{a}{b+c}\), then show that cos θ = \(\frac{2 \sqrt{b c}}{b+c} \cdot \cos \frac{A}{2}\)
Solution:

(ii) If a = (b + c) cos θ, then prove that sin θ = \(\frac{2 \sqrt{b c}}{b+c} \cos \left(\frac{A}{2}\right)\)
Solution:

(iii) For any anlge θ show that a cos θ = b cos (C + θ) + c cos (B – θ).
Solution:
b cos (C + θ) + c cos (B – θ)
= b (cos C . cos θ – sin C sin θ) + c (cos B cos θ + sin B sin θ)
= (b cos C + c cos B) cos θ + (-b sin C + C sin B) sin θ
= a cos θ + (-2R sin B sin C + 2R sin B sin C) sin θ
= a cos θ

Question 4.
If the angles of ∆ABC are in A.P and b : c = √3 : √2 , then show that A = 75°.
Solution:
∵ The angles A, B, C of a triangle are in A.P.
⇒ 2B = A + C
⇒ 3B = A + B + C
⇒ 3B = 180°
⇒ B = 60°

Question 5.
If \(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}=\frac{\sin C}{\sin (A-B)}\), prove that ∆ABC is either isosceles or right-angled.
Solution:

⇒ sin 2A = sin 2B
⇒ A = B
⇒ ∆ABC is isosceles
or 2A = 180° – 2B
or A = 90° – B
or A + B = 90°
so A ≠ B ⇒ ∆ABC is a right-angled triangle
∴ ∆ABC is either isosceles (or) right-angled.

Question 6.
If cos A + cos B + cos C = \(\frac{3}{2}\), then show that the triangle is equilateral.
Solution:

Question 7.
If cos² A + cos² B + cos² C = 1, then show that ∆ABC is right-angled.
Solution:
Given cos² A + cos² B + cos² C = 1 …….(1)
Now cos² A + cos² B + cos² C
= cos² A + 1 – sin² B + cos² C
= 1 + (cos² A – sin² B) + cos² C
= 1 + cos (A + B) cos (A – B) + cos² C
= 1 + cos (180° – C) cos (A – B) + cos² C
= 1 – cos C . cos (A – B) + cos² C
= 1 – cos C (cos (A – B) – cos C)
= 1 – cos C (cos (A – B) – cos (180° – \(\overline{A+B}\)))
= 1 – cos C (cos (A – B) + cos (A + B))
= 1 – cos C (2 cos A cos B)
= 1 – 2 cos A cos B cos C
Substituting in (1) we get
1 – 2 cos A cos B cos C = 1
∴ 2 cos A cos B cos C = 0
∴ cos A = 0 or cos B = 0 or cos C = 0
i.e., A = 90° or B = 90° or C = 90°
∴ ∆ABC is right-angled.

Question 8.
If a² + b² + c² = 8R², then prove that the triangle is right angled.
Solution:
Given a² + b² + c² = 8R²
⇒ 4R² (sin² A + sin² B + sin² C) = 8R²
⇒ sin² A + sin² B + sin² C = 2 ……(1)
Now sin² A + sin² B + sin² C
= 1 – cos² A + sin² B + sin² C
= 1 – (cos² A – sin² B) + sin² C
= 1 – cos (A + B) . cos (A – B) + sin² C
= 1 – cos (180° – C) cos (A – B) + sin² C
= 1 + cos C cos (A – B) + 1 – cos² C
= 2 + cos C (cos (A – B) – cos C)
= 2 + cos C (cos (A – B) – cos (180° – \(\overline{A+B}\)))
= 2 + cos C (cos (A – B) + cos (A + B))
= 2 + cos C (2 cos A cos B)
= 2 + 2 cos A cos B cos C
Substituting in (1), we get
2 + 2 cos A cos B cos C = 2
2 cos A cos B cos C = 0
⇒ cos A = 0 or cos B = 0 or cos C = 0
∴ A = 90° or B = 90° or C = 90°
∴ ΔABC is right-angled.

Question 9.
If cot \(\frac{A}{2}\), cot \(\frac{B}{2}\), cot \(\frac{C}{2}\) are in A.P., then prove that a, b, c are in A.P.
Solution:
∵ cot \(\frac{A}{2}\), cot \(\frac{B}{2}\), cot \(\frac{C}{2}\) are in A.P
⇒ \(\frac{(s)(s-a)}{\Delta}, \frac{(s)(s-b)}{\Delta}, \frac{(s)(s-c)}{\Delta}\) are in A.P.
⇒ s – a, s – b, s – c are in A.P
⇒ -a, -b, -c are in A.P
⇒ a, b, c are in A.P

Question 10.
If \(\sin ^{2} \frac{A}{2}, \sin ^{2} \frac{B}{2}, \sin ^{2} \frac{C}{2}\) are in H.P., show that a, b, c are in H.P.
Solution:

Question 11.
If C = 90° then prove that \(\frac{a^{2}+b^{2}}{a^{2}-b^{2}}\) sin (A – B) = 1
Solution:

Question 12.
Show that \(\frac{a^{2}}{4} \sin 2 C+\frac{c^{2}}{4} \sin 2 A\) = ∆
Solution:
LHS = \(\frac{a^{2}}{4} \sin 2 C+\frac{c^{2}}{4} \sin 2 A\)
= \(\frac{4 R^{2} \sin ^{2} A}{4}\) 2 sin C cos C + \(\frac{4 R^{2} \sin ^{2} C}{4}\) 2 sin A cos A
= 2R² sin²A sin C cos C + 2R² sin²C sin A cos A
= 2R² sin A sin C (sin A cos C + cos A sin C)
= 2R² sin A sin C sin(A + C)
= 2R² sin A sin C sin(180 – B)
= 2R² sin A sin B sin C
= ∆
= RHS

Question 13.
A lamp post is situated at the middle point I the side AC of a triangular plot A with BC = 7 meters, CA = 8 meters, and AB = 9 meters, lamp post subtends an angle of 15° at point B. find H height of the lamp post.
Solution:
Let MP be the height of the lamp post

Question 14.
Two ships leave a port at the same time. One goes 24 km. per hour in the direction N45°E and other travels 32 km per hour in the direction S75°E. Find the distance between the ships at the end of 3 hours.
Solution:
Given the first ship goes 24 km per hour After 3 hours the first ship goes 72 km.

Given the second ship goes 32 km per hour After 3 hours, the second ship goes 96 km.
Let AB = x
∠AOB = 180° – (75 + 45) = 60°
Apply cosine rule for ∆AOB,
cos 60° = \(\frac{(72)^{2}+(96)^{2}-x^{2}}{2(72)(96)}\)
⇒ \(\frac{1}{2}=\frac{5184+9216-x^{2}}{13824}\)
⇒ 13824 = 28800 – 2x²
⇒ 2x² = 14976
⇒ x² = 7488
⇒ x = 86.4 (Appx)
At the end of 3 hours the difference between ships was 86.4 km.

Question 15.
A tree stands vertically on the slant of the hill. From A point on the ground 35 meters down the hill from the base of the tree, the angle, the elevation of the top of the tree is 60° if the angle of elevation of the foot of the tree from A is 15°, then find the height of the tree.
Solution:
Let BC be the height of the tree
BC = h
Let BD = x, AD = y
Given AB = 35 m

Question 16.
The upper \(\frac{3}{4}\)th portion of a vertical pole subtends an angle tan-1(\(\frac{3}{5}\)) at a point in the horizontal plane through its foot and at a distance 40 m from the foot. Given that the vertical pole is at a height less than 100 m from the ground, find its height.
Solution:

⇒ 6400 + h² = 200h
⇒ h² – 200h + 6400 = 0
⇒ h² – 160h – 40h + 6400 = 0
⇒ h(h – 160) – 40(h – 160) = 0
⇒ (h – 160) (h – 40) = 0
⇒ h = 40 or 160
but the height of the pole should be less than 100m
∴ h = 40 m

Question 17.
AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7 m. From D, the angle of elevation of point A is 45°. Find the height of the pole.
Solution:
Let ‘h’ be the height of the pole
AB = h
given CD = 7 m
∠ACB = 60°, ∠ADB = 45°, Let BC = x

Question 18.
Let an object be placed at some height h cm and let P and Q two points of observation which are at a distance of 10 cm apart on a line inclined at an angle of 15° to the horizontal. If the angles of elevation of the object from P and Q are 30° and 60° respectively then find h.
Solution:
Let AB = Height of the object from ‘A’ = h m
Given that P & Q are two observations,
PQ = 10 cms

From ∆APB,
∠P = 30; ∠A = 90; ∠B = ?
A + P + B = 180°
⇒ 30° + 30° + B = 180°
⇒ B = 180° – 120°
⇒ B = 60°
From ∆BQC,
∠Q = 60°; ∠C = 90°; ∠B = ?
Q + C + B = 180°
⇒ 60° + 90° + B = 180°
⇒ B = 180° – 150°
⇒ B = 30°
From ∆BQP,
∠P = 15; ∠B = 30; ∠Q = ?
P + B + Q = 180°
⇒ 15° + 30° + Q = 180°
⇒ Q = 180° – 45°
⇒ Q = 135°
Applying the ‘sin’ rule for ∆BQP

From ∆PAB,
sin 30° = \(\frac{B A}{B P}\)
BP . sin 30° = AB = h
√2 × 10 × \(\frac{1}{2}\) = AB = h
5√2 = AB = h
∴ h = 5√2

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Hyperbolic Functions Solutions Exercise 9(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Hyperbolic Functions Solutions Exercise 9(a)

Question 1.
If sinh x = \(\frac{3}{4}\), find cosh (2x) and sinh (2x).
Solution:

Question 2.
If sinh x = 3, then show that x = log_e(3 + √10).
Solution:

Question 3.
Prove that
(i) tanh (x – y) = \(\frac{\tanh x-\tanh y}{1-\tanh x \tanh y}\)
Solution:

(ii) coth (x – y) = \(\frac{{coth} x \cdot {coth} y-1}{{coth} y-{coth} x}\)
Solution:

Question 4.
Prove that
(i) (cosh x – sinh x)ⁿ = cosh (nx) – sinh (nx), for any n ∈ R.
Solution:

(ii) (cosh x + sinh x)ⁿ = cosh (nx) + sinh (nx), for any n ∈ R.
Solution:

Question 5.
Prove that \(\frac{\tanh x}{{sech} x-1}+\frac{\tanh x}{{sech} x+1}\) = -2 cosech x, for x ≠ 0
Solution:

Question 6.
Prove that \(\frac{\cosh x}{1-\tanh x}+\frac{\sinh x}{1-{coth} x}\) = sinh x + cosh x, for x ≠ 0
Solution:

Question 7.
For any x ∈ R, prove that cosh⁴x – sinh⁴x = cosh (2x)
Solution:
L.H.S = cosh⁴x – sinh⁴x
= (cosh²x)² – (sinh²x)²
= [cosh²x – sinh²x] [cosh²x + sinh²x]
= (1) cosh (2x)
= cosh (2x)
∴ cosh⁴x – sinh⁴x = cosh (2x)

Question 8.
If u = \(\log _{e}\left(\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)\) and if cos θ > 0,then prove that cosh u = sec θ.
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(e) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(e)

I.

Question 1.
Find the adjoint and inverse of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & -3 \\
4 & 6
\end{array}\right]\)
Solution:

(ii) \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\)
Solution:

(iv) \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\)
Solution:

Question 2.
If A = \(\left[\begin{array}{cc}
a+i b & c+i d \\
-c+i d & a-i b
\end{array}\right]\), a² + b² + c² + d² = 1, then find the inverse of A.
Solution:

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
0 & -1 & 4 \\
-2 & 2 & 1
\end{array}\right]\), then find A^-1
Solution:

Question 4.
If A = \(\left|\begin{array}{ccc}
-1 & -2 & -2 \\
2 & 1 & -2 \\
2 & -2 & 1
\end{array}\right|\), then show that the adjoint of A = 3A’ find A^-1.
Solution:

Question 5.
If abc ≠ 0, find the inverse of \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right]\)
Solution:

II.

Question 1.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\) and B = \(\frac{1}{2}\left[\begin{array}{lll}
b+c & c-a & b-a \\
c-b & c+a & a-b \\
b-c & a-c & a+b
\end{array}\right]\) then show that ABA^-1 is a diagonal matrix.
Solution:

Question 2.
If 3A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
-2 & 2 & -1
\end{array}\right]\) then show that A^-1 = A’
Solution:

Question 3.
If A = \(\left[\begin{array}{rrr}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\), then show that A^-1 = A³
Solution:

∴ A⁴ = I
det A = 3(1) – 3(-2) + 4(-2) = 1
∵ A ≠ 0 ⇒ A^-1 exists
∵ A⁴ = I
Multiply with A^-1
A⁴ (A^-1) = I (A^-1)
⇒ A³ (AA^-1) = A^-1
⇒ A³ (I) = A^-1
∴ A^-1 = A³

Question 4.
If AB = I or BA = I, then prove that A is invertible and B = A^-1
Solution:
Given AB = I
⇒ AB| = |1|
⇒ |A| |B| = 1
⇒ |A| ≠ 0
∴ A is a non-singular matrix and BA = I
⇒ |BA| = |I|
⇒ |B| |A| = 1
⇒ |A| ≠ 0
∴ A is a non-singular matrix.
AB = I or BA = I, A is invertible.
∴ A^-1 exists.
AB = I
⇒ A^-1AB = A^-1I
⇒ IB = A^-1
⇒ B = A^-1
∴ B = A^-1

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(d) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(d)

I.

Question 1.
Find the determinants of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & 1 \\
1 & -5
\end{array}\right]\)
Solution:
det A = ad – bc
= 2(-5) – 1(1)
= -10 – 1
= -11

(ii) \(\left[\begin{array}{cc}
4 & 5 \\
-6 & 2
\end{array}\right]\)
Solution:
det A = 4(2) – (-6)(5)
= 8 + 30
= 38

(iii) \(\left[\begin{array}{cc}
\mathbf{i} & 0 \\
0 & -\mathbf{i}
\end{array}\right]\)
Solution:
det A = -i² – 0
= 1 – 0
= 1

(iv) \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\)
Solution:
det A = 0(0 – 1) – 1(0 – 1) + 1(1 – 0)
= 1 + 1
= 2

(v) \(\left[\begin{array}{ccc}
1 & 4 & 2 \\
2 & -1 & 4 \\
-3 & 7 & 6
\end{array}\right]\)
Solution:
det A = 1(-6 – 28) – 4(12 + 12) + 2(14 – 3)
= -34 – 96 + 22
= -108

(vi) \(\left[\begin{array}{ccc}
2 & -1 & 4 \\
4 & -3 & 1 \\
1 & 2 & 1
\end{array}\right]\)
Solution:
det A = 2(-3 – 2) + 1(4 – 1) + 4(8 + 3)
= -10 + 3 + 44
= 37

(vii) \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
4 & -1 & 7 \\
2 & 4 & -6
\end{array}\right]\)
Solution:
det A = 0 since R₁ and R₃ are proportional.

(viii) \(\left[\begin{array}{lll}
a & h & g \\
\text { h } & b & f \\
g & f & c
\end{array}\right]\)
Solution:
det A = a(bc – f²) – h(ch – fg) + g(hf – bg)
= abc – af² – ch² + fgh + fgh – bg²
= abc + 2fgh – af² – bg² – ch²

(ix) \(\left[\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{b} & \mathbf{c} & \mathbf{a} \\
\mathbf{c} & \mathbf{a} & \mathbf{b}
\end{array}\right]\)
Solution:
det A = a(bc – a²) – b(b² – ac) + c(ab – c²)
= abc – a³ – b³ + abc + abc – c³
= 3abc – a³ – b³ – c³

(x) \(\left[\begin{array}{ccc}
1^{2} & 2^{2} & 3^{2} \\
2^{2} & 3^{2} & 4^{2} \\
3^{2} & 4^{2} & 5^{2}
\end{array}\right]\)
Solution:
det A = \(\left|\begin{array}{ccc}
1 & 4 & 9 \\
4 & 9 & 16 \\
9 & 16 & 25
\end{array}\right|\)
= 1(225 – 256) – 4(100 – 144) – 9(64 – 81)
= -31 + 176 – 153
= -184 + 176
= -8

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & x
\end{array}\right]\) and det A = 45 then find x.
Solution:
det A = 45
\(\left|\begin{array}{ccc}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & x
\end{array}\right|\) = 45
⇒ 3x + 24 = 45
⇒ 3x – 45 + 24 = 0
⇒ 3x – 21 = 0
⇒ x = 7

II.

Question 1.
Show that \(\left|\begin{array}{lll}
b c & b+c & 1 \\
c a & c+a & 1 \\
a b & a+b & 1
\end{array}\right|\) = (a – b)(b – c)(c – a)
Solution:

Question 2.
Show that \(\left|\begin{array}{ccc}
\mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} \\
\mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a} \\
\mathbf{a} & \mathbf{b} & \mathbf{c}
\end{array}\right|\) = a³ + b³ + c³ – 3abc
Solution:

= (a + b + c) [(-ac + b²) – (-c² + ab) + (-bc + a²)]
= (a + b + c) (-ac + b² + c² – ab – bc + a²)
= (a + b + c) (a² + b² + c² – ab – bc – ca)
= a³ + b³ + c³ – 3abc

Question 3.
Show that \(\left|\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & \mathbf{z} & \mathbf{x}+\mathbf{y}
\end{array}\right|\) = 4xyz
Solution:
L.H.S = \(\left|\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & \mathbf{z} & \mathbf{x}+\mathbf{y}
\end{array}\right|\)
= (y + z) [(z + x) (x + y) – yz] – x[y(x + y) – yz] + x[yz – z(z + x)]
= (y + z) (zx + yz + x² + xy – yz) – x(xy + y² – yz) + x(yz – z² – zx)
= (y + z) (zx + x² + xy) – x(xy + y² – yz) + x(yz – z² – zx)
= xyz + x²y + xy² + xz² + x²z + xyz – x²y – xy² + xyz + xyz – xz² – x²z
= 4xyz
= R.H.S

Question 4.
If \(\left|\begin{array}{ccc}
a & a^{2} & 1+a^{3} \\
b & b^{2} & 1+b^{3} \\
c & c^{2} & 1+c^{3}
\end{array}\right|\) = 0 and \(\left|\begin{array}{ccc}
a & a^{2} & 1 \\
b & b^{2} & 1 \\
c & c^{2} & 1
\end{array}\right|\) ≠ 0 then show that abc = -1
Hint: If each element in a row (column) of a square matrix is the sum of two numbers, then its discriminant can be expressed as the sum of discriminants of two square matrices.
Solution:

Question 5.
Without expanding the determinant, prove that
(i) \(\left|\begin{array}{ccc}
a & a^{2} & b c \\
b & b^{2} & c a \\
c & c^{2} & a b
\end{array}\right|=\left|\begin{array}{ccc}
1 & a^{2} & a^{3} \\
1 & b^{2} & b^{3} \\
1 & c^{2} & c^{3}
\end{array}\right|\)
Solution:

(ii) \(\left|\begin{array}{ccc}
a x & b y & c z \\
x^{2} & y^{2} & z^{2} \\
1 & 1 & 1
\end{array}\right|=\left|\begin{array}{ccc}
a & b & c \\
x & y & z \\
y z & z x & x y
\end{array}\right|\)
Solution:

(iii) \(\left|\begin{array}{lll}
1 & b c & b+c \\
1 & c a & c+a \\
1 & a b & a+b
\end{array}\right|=\left|\begin{array}{ccc}
1 & a & a^{2} \\
1 & b & b^{2} \\
1 & c & c^{2}
\end{array}\right|\)
Solution:
L.H.S = \(\left|\begin{array}{ccc}
1 & b c & b+c \\
1 & c a & c+a \\
1 & a b & a+b
\end{array}\right|\)

= (b – a) (c – a) (c + a – b – a)
= (a – b) (b – c) (c – a)
∴ LHS = RHS

Question 6.
If ∆₁ = \(\left|\begin{array}{ccc}
a_{1}^{2}+b_{1}+c_{1} & a_{1} a_{2}+b_{2}+c_{2} & a_{1} a_{3}+b_{3}+c_{3} \\
b_{1} b_{2}+c_{1} & b_{2}^{2}+c_{2} & b_{2} b_{3}+c_{3} \\
c_{3} c_{1} & c_{3} c_{2} & c_{3}^{2}
\end{array}\right|\) and ∆₂ = \(\left|\begin{array}{lll}
a_{1} & b_{2} & c_{2} \\
a_{2} & b_{2} & c_{2} \\
a_{3} & b_{3} & c_{3}
\end{array}\right|\), then find the value of \(\frac{\Delta_{1}}{\Delta_{2}}\)
Solution:

Question 7.
If ∆₁ = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \alpha & 1
\end{array}\right|\), ∆₂ = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\) and ∆₁ = ∆₂, then show that cos²α + cos²β + cos²γ = 1
Solution:
∆₁ = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \alpha & 1
\end{array}\right|\)
= 1(1 – cos²γ) – cos α (cos α – cos β cos γ) + cos β (cos α cos γ – cos β)
= 1 – cos²γ – cos²α + cos α cos β cos γ + cos α cos β cos γ – cos²β
= 1 – cos²γ – cos²α – cos²β + 2 cos α cos β cos γ
∆₂ = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\)
= 0(0 – cos²γ) – cos α (0 – cos γ cos β) + cos β (cos α cos γ – 0)
= cos α cos β cos γ + cos α cos β cos γ
= 2 cos α cos β cos γ
Given ∆₁ = ∆₂
1 – cos²α – cos²β – cos²γ + 2 cos α cos β cos γ = 2 cos α cos β cos γ
1 – cos²α – cos²β – cos²γ = 0
1 = cos²α + cos²β + cos²γ

III.

Question 1.
Show that \(\left|\begin{array}{ccc}
\mathbf{a}+\mathbf{b}+2 \mathbf{c} & \mathbf{a} & \mathbf{b} \\
\mathbf{c} & \mathbf{b}+\mathbf{c}+\mathbf{2} \mathbf{a} & \mathbf{b} \\
\mathbf{c} & \mathbf{a} & \mathbf{c}+\mathbf{a}+\mathbf{2} \mathbf{b}
\end{array}\right|\) = 2(a + b + c)³
Solution:

Question 2.
Show that \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|^{2}\) = \(\left|\begin{array}{ccc}
2 b c-a^{2} & c^{2} & b^{2} \\
c^{2} & 2 a c-b^{2} & a^{2} \\
b^{2} & a^{2} & 2 a b-c^{2}
\end{array}\right|\) = (a³ + b³ + c³ – 3abc)²
Solution:

Question 3.
Show that \(\left|\begin{array}{ccc}
a^{2}+2 a & 2 a+1 & 1 \\
2 a+1 & a+2 & 1 \\
3 & 3 & 1
\end{array}\right|\) = (a – 1)³
Solution:

Question 4.
Show that \(\left|\begin{array}{ccc}
a & b & c \\
a^{2} & b^{2} & c^{2} \\
a^{3} & b^{3} & c^{3}
\end{array}\right|\) = abc(a – b)(b – c)(c – a)
Solution:

Question 5.
Show that \(\left|\begin{array}{ccc}
-2 \mathbf{a} & \mathbf{a}+\mathbf{b} & \mathbf{c}+\mathbf{a} \\
\mathbf{a}+\mathbf{b} & -\mathbf{2} \mathbf{b} & \mathbf{b}+\mathbf{c} \\
\mathbf{c}+\mathbf{a} & \mathbf{c}+\mathbf{b} & -2 \mathbf{c}
\end{array}\right|\) = 4(a + b) (b + c) (c + a)
Solution:

∴ (c + a) is a factor for ∆
Similarly a + b, b + c are also factors ∆.
∵ ∆ is a third-degree expression in a, b, c.
∆ = k(a + b) (b + c) (c + a),
where k is a non-zero scalar.
Put a = 1, b = 1, c = 1, then
\(\left|\begin{array}{ccc}
-2 & 2 & 2 \\
2 & -2 & 2 \\
2 & 2 & -2
\end{array}\right|\) = k(1 + 1) (1 + 1) (1 + 1)
⇒ -2(4 – 4) – 2(-4 – 4) + 2(4 + 4) = 8k
⇒ 16 + 16 = 8k
⇒ k = 4
∴ ∆ = 4(a + b) (b + c) (c + a)
Hence \(\left|\begin{array}{ccc}
-2 \mathbf{a} & \mathbf{a}+\mathbf{b} & \mathbf{c}+\mathbf{a} \\
\mathbf{a}+\mathbf{b} & -\mathbf{2} \mathbf{b} & \mathbf{b}+\mathbf{c} \\
\mathbf{c}+\mathbf{a} & \mathbf{c}+\mathbf{b} & -2 \mathbf{c}
\end{array}\right|\) = 4(a + b) (b + c) (c + a)

Question 6.
Show that \(\left|\begin{array}{lll}
\mathbf{a}-\mathbf{b} & \mathbf{b}-\mathbf{c} & \mathbf{c}-\mathbf{a} \\
\mathbf{b}-\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{c}-\mathbf{a} & \mathbf{a}-\mathbf{b} & \mathbf{b}-\mathbf{c}
\end{array}\right|\)
Solution:
L.H.S = \(\left|\begin{array}{ccc}
0 & 0 & 0 \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\) = 0
By R₁ → R₁ + (R₂ + R₃)

Question 7.
Show that \(\left|\begin{array}{ccc}
1 & a & a^{2}-b c \\
1 & b & b^{2}-c a \\
1 & c & c^{2}-a b
\end{array}\right|\) = 0
Solution:

Question 8.
Show that \(\left|\begin{array}{lll}
\mathbf{x} & \mathbf{a} & \mathbf{a} \\
\mathbf{a} & \mathbf{x} & \mathbf{a} \\
\mathbf{a} & \mathbf{a} & \mathbf{x}
\end{array}\right|\) = (x + 2a) (x – a)²
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(c) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(c)

I.

Question 1.
If A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
-1 & 1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & 1 & -2
\end{array}\right]\), then find (AB’)’.
Solution:

Question 2.
If A = \(\left[\begin{array}{cc}
-2 & 1 \\
5 & 0 \\
-1 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-2 & 3 & 1 \\
4 & 0 & 2
\end{array}\right]\) then find 2A + B’ and 3B’ – A.
Solution:

Question 3.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
-5 & 3
\end{array}\right]\), then find A + A’ and A.A’
Solution:

Question 4.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
2 & 5 & 6 \\
3 & x & 7
\end{array}\right]\) is a symmetric matrix, then find x.
Hint: ‘A’ is a symmetric matrix ⇒ A^T = A
Solution:
A is a symmetric matrix
⇒ A’ = A

Equating 2nd row, 3rd column elements we get x = 6.

Question 5.
If A = \(\left[\begin{array}{ccc}
0 & 2 & 1 \\
-2 & 0 & -2 \\
-1 & x & 0
\end{array}\right]\) is a skew-symmetric matrix, find x.
Solution:
∵ A is a skew-symmetric matrix
⇒ A^T = -A

Equating second-row third column elements we get x = 2.

Question 6.
Is \(\left[\begin{array}{ccc}
0 & 1 & 4 \\
-1 & 0 & 7 \\
-4 & -7 & 0
\end{array}\right]\) symmetric or skewsymmetric?
Solution:

∴ A is a skew-symmetric matrix.

II.

Question 1.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that A.A’ = A’. A = I₂
Solution:

From (1), (2) we get A.A’ = A’. A = I₂

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & 5 & 3 \\
2 & 4 & 0 \\
3 & -1 & -5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
2 & -1 & 0 \\
0 & -2 & 5 \\
1 & 2 & 0
\end{array}\right]\) then find 3A – 4B’.
Solution:

Question 3.
If A = \(\left[\begin{array}{cc}
7 & -2 \\
-1 & 2 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-2 & -1 \\
4 & 2 \\
-1 & 0
\end{array}\right]\) then find AB’ and BA’.
Solution:

Question 4.
For any square matrix A, Show that AA’ is symmetric.
Solution:
A is a square matrix
(AA’)’ = (A’)’A’ = A.A’
∵ (AA’)’ = AA’
⇒ AA’ is a symmetric matrix.

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(b)

I.

Question 1.
Find the following products wherever possible.
Hint: (1 × 3) by (3 × 1) = 1 × 1
(i) \(\left[\begin{array}{lll}
-1 & 4 & 2
\end{array}\right]\left[\begin{array}{l}
5 \\
1 \\
3
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
2 & 1 & 4 \\
6 & -2 & 3
\end{array}\right]\left[\begin{array}{l}
1 \\
2 \\
1
\end{array}\right]\)
(iii) \(\left[\begin{array}{cc}
3 & -2 \\
1 & 6
\end{array}\right]\left[\begin{array}{cc}
4 & -1 \\
2 & 5
\end{array}\right]\)
(iv) \(\left[\begin{array}{lll}
2 & 2 & 1 \\
1 & 0 & 2 \\
2 & 1 & 2
\end{array}\right]\left[\begin{array}{ccc}
-2 & -3 & 4 \\
2 & 2 & -3 \\
1 & 2 & -2
\end{array}\right]\)
(v) \(\left[\begin{array}{ccc}
3 & 4 & 9 \\
0 & -1 & 5 \\
2 & 6 & 12
\end{array}\right]\left[\begin{array}{ccc}
13 & -2 & 0 \\
0 & 4 & 1
\end{array}\right]\)
(vi) \(\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]\left[\begin{array}{ccc}
2 & 1 & 4 \\
6 & -2 & 3
\end{array}\right]\)
(vii) \(\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
(viii) \(\left[\begin{array}{ccc}
0 & c & -b \\
-c & 0 & a \\
b & -a & 0
\end{array}\right]\left[\begin{array}{ccc}
a^{2} & a b & a c \\
a b & b^{2} & b c \\
a c & b c & c^{2}
\end{array}\right]\)
Solution:

(v) \(\left[\begin{array}{ccc}
3 & 4 & 9 \\
0 & -1 & 5 \\
2 & 6 & 12
\end{array}\right]\left[\begin{array}{ccc}
13 & -2 & 0 \\
0 & 4 & 1
\end{array}\right]\)
First matrix is a 3 × 3 matrix and second matrix is 2 × 3 matrix.
No. of columns in the first matrix ≠ No. of rows in the second matrix.
∴ Matrix product is not possible.

(vi) \(\left[\begin{array}{c}
1 \\
-2 \\
1
\end{array}\right]\left[\begin{array}{ccc}
2 & 1 & 4 \\
6 & -2 & 3
\end{array}\right]\)
No. of columns in first matrix = 1
No. of rows in second matrix = 2
No. of columns in the first matrix ≠ No. of rows in the second matrix
Multiplication of matrices is not possible.

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\), do AB and BA exist? If they exist, find them. Do A and B commute with respect to multiplication?
Solution:

AB ≠ BA
∴ A and B are not commutative with respect to the multiplication of matrices.

Question 3.
Find A² where A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\)
Solution:

Question 4.
If A = \(\left[\begin{array}{ll}
i & 0 \\
0 & i
\end{array}\right]\), find A².
Solution:

Question 5.
If A = \(\left[\begin{array}{cc}
i & 0 \\
0 & -i
\end{array}\right]\), B = \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
0 & \mathbf{i} \\
\mathbf{i} & \mathbf{0}
\end{array}\right]\), and I is the unit matrix of order 2, then show that
(i) A² = B² = C² = -I
(ii) AB = -BA = -C
Solution:

Question 6.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\), find AB. Find BA if it exists.
Solution:
Given A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\)

The order of AB is 2 × 3
BA does not exist since no. of columns in B ≠ No. of rows in A.

Question 7.
If A = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & k
\end{array}\right]\) and A² = 0, then find the value of k.
Solution:

II.

Question 1.
If A = \(\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right]\) then find A⁴.
Solution:

Question 2.
If A = \(\left[\begin{array}{ccc}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{array}\right]\) then find A³.
Solution:

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & -1 \\
3 & -1 & 1
\end{array}\right]\), then find A³ – 3A² – A – 3I, where I is unit matrix of order 3 × 3.
Solution:

Question 4.
If I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) and E = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\), show that (aI + bE)³ = a³I + 3a²bE, Where I is unit matrix of order 2.
Solution:

III.

Question 1.
If A = [a₁, a₂, a₃,], then for any integer n ≥ 1 show that Aⁿ = \(\left[\begin{array}{lll}
a_{1}, & a_{2}^{n}, & a_{3}^{n}
\end{array}\right]\)
Solution:
Given A = diag [a₁, a₂, a₃,] = \(\left[\begin{array}{ccc}
a_{1} & 0 & 0 \\
0 & a_{2} & 0 \\
0 & 0 & a_{3}
\end{array}\right]\)
Aⁿ = diag \(\left[\begin{array}{lll}
a_{1}^{n} & a_{2}^{n} & a_{3}^{n}
\end{array}\right]=\left[\begin{array}{ccc}
a_{1}^{n} & 0 & 0 \\
0 & a_{2}^{n} & 0 \\
0 & 0 & a_{3}^{n}
\end{array}\right]\)
This problem can be should by using Mathematical Induction
put n = 1
A¹ = \(\left[\begin{array}{ccc}
a_{1} & 0 & 0 \\
0 & a_{2} & 0 \\
0 & 0 & a_{3}
\end{array}\right]\)
∴ The result is true for n = 1
Assume the result is true for n = k
A^k = \(\left[\begin{array}{ccc}
a_{1}^{k} & 0 & 0 \\
0 & a_{2}^{k} & 0 \\
0 & 0 & a_{3}^{k}
\end{array}\right]\)
Consider

∴ The result is true for n = k + 1
Hence by the Principle of Mathematical Induction, the statement is true ∀ n ∈ N

Question 2.
If θ – φ = \(\frac{\pi}{2}\), then show that \(\left[\begin{array}{cc}
\cos ^{2} \theta & \cos \theta \sin \theta \\
\cos \theta \sin \theta & \sin ^{2} \dot{\theta}
\end{array}\right]\) \(\left[\begin{array}{cc}
\cos ^{2} \phi & \cos \phi \sin \phi \\
\cos \phi \sin \phi & \sin ^{2} \phi
\end{array}\right]\) = 0
Solution:

Question 3.
If A = \(\left[\begin{array}{rr}
3 & -4 \\
1 & -1
\end{array}\right]\) then show that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), for any integer n ≥ 1 by using Mathematical Induction.
Solution:
We shall prove the result by Mathematical Induction.

∴ The given result is true for n = k + 1
By Mathematical Induction, the given result is true for all positive integral values of n.

Question 4.
Give examples of two square matrices A and B of the same order for which AB = 0 but BA ≠ 0.
Solution:

Question 5.
A Trust fund has to invest ₹ 30,000 in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds if the trust fund must obtain an annual total interest of (a) ₹ 1800 (b) ₹ 2000
Solution:
Let the first bond be ‘x’ and the second bond be 30,000 – x respectively
The rate of interest is 0.05 and 0.07 respectively.
(a) \([x, 30,000-x]\left[\begin{array}{l}
0.05 \\
0.07
\end{array}\right] \quad=[1800]\)
[0.05x + 0.07(30,000 – x)] = 1800
\(\frac{5}{100} x+\frac{7}{100}(30,000-x)=1800\)
5x + 21,0000 – 7x = 1,80,000
-2x = 1,80,000 – 2,10,000 = -30,000
x = 15,000
∴ First bond = 15,000
Second bond = 30,000 – 15,000 = 15,000

(b) \(\left[\begin{array}{ll}
x & 30,000-x
\end{array}\right]\left[\begin{array}{l}
0.05 \\
0.07
\end{array}\right]=[2000]\)
[0.05x + 0.07(30,000 – x)] = [2000]
\(\frac{5 x}{100} \times \frac{7}{100}(30,000-x)=2000\)
5x + 2,10,000 – 7x = 2,00,000
-2x = 2,00,000 – 2,10,000
-2x = -10,000
x = 5,000
∴ First bond = 5000
Second bond = 30,000 – 5000 = 25,000

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(a)

I.

Question 1.
Write the following as a single matrix.
(i) \(\left[\begin{array}{lll}
2 & 1 & 3
\end{array}\right]+\left[\begin{array}{lll}
1 & 0 & 0
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
3 & 9 & 0 \\
1 & 8 & -2
\end{array}\right]+\left[\begin{array}{ccc}
4 & 0 & 2 \\
7 & 1 & 4
\end{array}\right]\)
(iii) \(\left[\begin{array}{c}
0 \\
1 \\
-1
\end{array}\right]+\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]\)
(iv) \(\left[\begin{array}{cc}
-1 & 2 \\
2 & -2 \\
3 & 1
\end{array}\right]-\left[\begin{array}{cc}
0 & 1 \\
-1 & 0 \\
-2 & 1
\end{array}\right]\)
Solution:

Question 2.
If A = \(\left[\begin{array}{cc}
-1 & 3 \\
4 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 1 \\
3 & -5
\end{array}\right]\), X = \(\left[\begin{array}{ll}
x_{1} & x_{2} \\
x_{3} & x_{4}
\end{array}\right]\) and A + B = X, then find the values of x₁, x₂, x₃ and x₄.
Solution:
A + B = X

∴ x₁ = 1, x₂ = 4, x₃ = 7, x₄ = -3

Question 3.
If A = \(\left[\begin{array}{ccc}
-1 & -2 & 3 \\
1 & 2 & 4 \\
2 & -1 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -2 & 5 \\
0 & -2 & 2 \\
1 & 2 & -3
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
-2 & 1 & 2 \\
1 & 1 & 2 \\
2 & 0 & 1
\end{array}\right]\) then find A + B + C.
Solution:

Question 4.
If A = \(\left[\begin{array}{ccc}
3 & 2 & -1 \\
2 & -2 & 0 \\
1 & 3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & -1 & 0 \\
2 & 1 & 3 \\
4 & -1 & 2
\end{array}\right]\) and X = A + B then find X.
Solution:

Question 5.
If \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]=\left[\begin{array}{cc}
5 & 2 \\
-2 & a-4
\end{array}\right]\) then find the values of x, y, z and a.
Solution:
Given \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]=\left[\begin{array}{cc}
5 & 2 \\
-2 & a-4
\end{array}\right]\)
∴ x – 3 = 5 ⇒ x = 3 + 5 = 8
2y – 8 = 2 ⇒ 2y = 8 + 2 = 10 ⇒ y = 5
z + 2 = -2 ⇒ z = -2 – 2 = -4
a – 4 = 6 ⇒ a = 4 + 6 = 10

II.

Question 1.
If \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]=\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\) then find the values of x, y, z and a.
Solution:
Given \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]=\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\)
∴ x – 1 = 1 ⇒ x = 1 + 1 = 2
5 – y = 3 ⇒ y = 5 – 3 = 2
z – 1 = 4 ⇒ z = 4 + 1 = 5
a – 5 = 0 ⇒ a = 5

Question 2.
Find the trace of \(\left[\begin{array}{ccc}
1 & 3 & -5 \\
2 & -1 & 5 \\
1 & 0 & 1
\end{array}\right]\)
Solution:
Trace of \(\left[\begin{array}{ccc}
1 & 3 & -5 \\
2 & -1 & 5 \\
1 & 0 & 1
\end{array}\right]\) = Sum of the diagonal elements
= 1 – 1 + 1
= 1

Question 3.
If A = \(\left[\begin{array}{rrr}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & -6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) find B – A and 4A – 5B.
Solution:

Question 4.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\) find 3B – 2A.
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Mathematical Induction Solutions Exercise 2(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Mathematical Induction Solutions Exercise 2(a)

Using mathematical induction, prove each of the following statements for all n ∈ N.

Question 1.
1² + 2² + 3² + …… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Solution:
Let p(n) be the given statement:
1² + 2² + 3² + ….. + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Since 1² = \(\frac{(1)(1+1)(2 \times 1+1)}{6}\)
⇒ 1 = 1 the formula is true for n = 1
i.e., p(1) is true.
Assume the statement p(n) is true for n = k
i.e., 1² + 2² + 3² + …… + 1k² = \(\frac{k(k+1)(2 k+1)}{6}\)
We show that the formula is true for n = k + 1
i.e., We show that p(k + 1) = \(\frac{(k+1)(k+2)(2 k+3)}{6}\)
(Where p(k) = 1² + 2² + 3² + … + k²)
We observe that
p(k + 1) = 1² + 2² + 3² + …… + (k)² + (k + 1)² = p(k) + (k + 1)²
Since p(k) = \(\frac{k(k+1)(2 k+1)}{6}\)
We have p(k + 1) = p(k) + (k + 1)²

∴ The formula holds for n = k + 1
∴ By the principle of mathematical induction, p(n) is true for all n ∈ N
i.e., the formula 1² + 2² + 3² + ……. + n² = \(\frac{n(n+1)(2 n+1)}{6}\) for all n ∈ N

Question 2.
2.3 + 3.4 + 4.5 + …… up to n terms = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)
Solution:
The nth term in the given series is (n + 1) (n + 2)
Let p(n) be the statement:
2.3 + 3.4 + 4.5 + …… + (n + 1) (n + 2) = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)
and let S(n) be the sum on the left-hand side.
Since S(1) = 2.3 = \(\frac{(1)(1+6+11)}{3}\) = 6
∴ The statement is true for n = 1
Assume that the statement p(n) is true for n = k
i.e., S(k) = 2.3 + 3.4 + …… + (k + 1) (k + 2) = \(\frac{k\left(k^{2}+6 k+11\right)}{3}\)
We show that the statement is true for n = k + 1
i.e., We show that S(k + 1) = \((k+1)\left[\frac{(k+1)^{2}+6(k+1)+11}{3}\right]\)
We observe that
S(k + 1) = 2.3 + 3.4 + 4.5 + + (k + 1) (k + 2) + (k + 2) (k + 3)
= S(k) + (k + 2) (k + 3)

∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
i.e., 2.3 + 3.4 + 4.5 + ……. + (n + 1) (n + 2) = \(\frac{n\left(n^{2}+6 n+11\right)}{3}\)

Question 3.
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)
Solution:
Let p(n) be the statement:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)
and let S(n) be the sum on the L.H.S.
Since S(1) = \(\frac{1}{1.3}=\frac{1}{1(2+1)}=\frac{1}{1.3}\)
∴ P(1) is true.
Assume that the statement p(n) is true for n = k
i.e., S(k) = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 k-1)(2 k+1)}\) = \(\frac{k}{2 k+1}\)
We show that the statement p(n) is true for n = k + 1
i.e., we show that s(k + 1) = \(\frac{k+1}{2(k+1)+1}\)
We observe that

∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
i.e., \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\)

Question 4.
4³ + 8³ + 12³ + …… up to n terms = 16n²(n + 1)².
Solution:
4, 8, 12,….. are in A.P., whose nth term is (4n)
Let p(n) be the statement:
4³ + 8³ + 12³ + ………. + (4n)³ = 16n²(n + 1)²
and S(n) be the sum on the L.H.S.
S(1) = 4³ = 16(1²) (1 + 1)² = 16(4) = 64 = 4³
∴ p(1) is true
Assume that the statement p(n) is true for n = k
i.e., S(k) = 4³ + 8³ + (12)³ + …… + (4k)³ = 16k²(k + 1)²
We show that the statement is true for n = k + 1
i.e., We show that S(k + 1) = 16(k + 1)² (k + 2)²
We observe that
S(k + 1) = 4³ + 8³ + 12³ + …… + (4k)³ + [4(k + 1)]³
= S(k) + [4(k + 1)]³
= 16k² (k + 1)² + 4³ (k + 1)³
= 16(k + 1)² [k² + 4(k + 1)]
= 16(k + 1)² [k² + 4k + 4]
= 16(k + 1)² (k + 2)²
= 16(k + 1)² \((\overline{k+1}+1)^{2}\)
∴ The formula holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e.,) 4³ + 8³ + 12³ + …… + (4n)³ = 16n²(n + 1)²

Question 5.
a + (a + d) + (a + 2d) + ……. up to n terms = \(\frac{n}{2}\) [2a + (n – 1)d]
Solution:
Let p(n) be the statement:
a + (a + d) + (a + 2d) + …… + [a + (n – 1)d] = \(\frac{n}{2}\) [2a + (n – 1)d]
and let the sum on the L.H.S. is denoted by S(n)
Since S(1) = a = \(\frac{1}{2}\) [2a + (1 – 1)d] = a
∴ p(1) is true.
Assume that the statement is true for n = k
(i.e.,) S(k) = a + (a + d) + (a + 2d) + ……. + [a + (k – 1)d] = \(\frac{k}{2}\) [2a + (k – 1 )d]
We show that the statement is true for n = k + 1
(i.e.,) we show that S(k + 1) = \(\left(\frac{k+1}{2}\right)[2 a+k d]\)
We observe that
S(k + 1) = a + (a + d) + (a + 2d) + …… + [a + (k – 1)d] + (a + kd)
= S(k) + (a + kd)
= \(\frac{k}{2}\) [2a + (k – 1)d] + (a + kd)
= \(\frac{k[2 a+(k-1) d]+2(a+k d)}{2}\)
= \(\frac{1}{2}\) [2ak + k(k – 1)d + 2a + 2kd]
= \(\frac{1}{2}\) [2a(k + 1) + k(k – 1 + 2)d]
= \(\frac{1}{2}\) (k + 1)(2a + kd)
∴ The statement holds for n = k + 1
∴ By the principle of mathematical inductions,
p(n) is true for all n ∈ N
(i.e.,) a + (a + d) + (a + 2d) + …… + [a + (n – 1)d] = \(\frac{n}{2}\) [2a + (n – 1)d]

Question 6.
a + ar + ar² + ……… up to n terms = \(\frac{a\left(r^{n}-1\right)}{r-1}\); r ≠ 1
Solution:
Let p(n) be the statement:
a + ar + a.r² + …… + a. r^n-1 = \(\frac{a\left(r^{n}-1\right)}{r-1}\), r ≠ 1
and let S(n) be the sum on the L.H.S
Since S(1) = a = \(\frac{a\left(r^{1}-1\right)}{r-1}\) = a
∴ p(1) is true
Assume that the statement is true for n = k
(i.e) S(k) = a + ar + ar² + ……… + a . r^k-1 = \(\frac{a\left(r^{k}-1\right)}{r-1}\)
We show that the statement is true for n = k + 1
(i.e) S(k + 1) = \(\frac{a\left(r^{k+1}-1\right)}{r-1}\)
Now S(k + 1) = a + ar + ar² + ……. + a r^k-1 + ar^k
= S(k) + a . r^k

∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e) a + ar + ar² + ……. + a.r^n-1 = \(\frac{a\left(r^{n}-1\right)}{r-1}\), r ≠ 1

Question 7.
2 + 7 + 12 + ……. + (5n – 3) = \(\frac{n(5 n-1)}{2}\)
Solution:
Let p(n) be the statement:
2 + 7 + 12 + ……. + (5n – 3) = \(\frac{n(5 n-1)}{2}\)
and let S(n) be the sum on the L.H.S
Since S(1) = 2 = \(\frac{1(5 \times 1-1)}{2}=\frac{4}{2}\) = 2
∴ p(1) is true
Assume that the statement is true for n = k
(i.e) S(k) = 2 + 7 + 12 + …….. + (5k – 3) = \(\frac{k(5 k-1)}{2}\)
We have to show that S(k + 1) = \(\frac{(k+1)(5 k+4)}{2}\)
We observe that S(k + 1) = 2 + 7 + 12 + ……. + (5k – 3) + (5k + 2)
= S(k) + (5k + 2)
= \(\frac{k(5 k-1)}{2}\) + (5k + 2)
= \(\frac{5 k^{2}-k+2(5 k+2)}{2}\)
= \(\frac{1}{2}\) [5k² + 9k + 4]
= \(\frac{1}{2}\) (k + 1) (5k + 4)
= \(\frac{1}{2}\) (k + 1) [5(k + 1) – 1]
∴ p(k + 1) is true
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N.
(i.e.,) 2 + 7 + 12 + …… + (5n – 3) = \(\frac{n(5 n-1)}{2}\)

Question 8.
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\frac{2 n+1}{n^{2}}\right)\) = (n + 1)²
Solution:
Let p(n) be the statement:
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots \ldots\left(1+\frac{2 n+1}{n^{2}}\right)\) = (n + 1)²
and let S(n) be the product on the LHS
since S(1) = 1 + 3 = 4 = (1 + 1)² = 4
∴ P(a) 4 time for n = 1
Assume that p(n) is true for n = k

= (k + 1)² + 2k + 3
= k² + 2k + 1 + 2k + 3
= k² + 4k + 4
= (k + 2)²
= (k + 1 + 1)²
∴ P(n) is true for n = k + 1
By the principle of Mathematical Induction,
p(n) is true & n ∈ N

Question 9.
(2n + 7) < (n + 3)²
Solution:
Let p (n) be the statement
When n = 1, 9 < 16
∴ p(n) is true for n = 1
Assume p (n) is true for n = k
(2k + 7) < (k + 3)²
We show that p(n) is true for n = k + 1
2(k + 1) + 7 = 2k + 7 + 2
< (k + 3)² + 2
< k² + 6k + 9 + 2 + 2k + 5 – 2k – 5
< (k + 4)² – (2k + 5)
< (k + 4)²
< (k + 1 + 3)²
∴ p(n) is true for n = k + 1
By the principle of Mathematical Induction
p(n) is true ∀ n ∈ N

Question 10.
1² + 2² + …… + n² > \(\frac{n^{3}}{3}\)
Solution:
Let P(n) by the statement
when n = 1, 1 > \(\frac{1}{3}\)
∴ p(n) is true for n = 1
Assume p (n) is true for n = k
1² + 2² + …… + k² > \(\frac{k^{3}}{3}\)
We show that p(n) is true for n = k + 1

∴ p(n) is true for n = k + 1
By the principle of Mathematical Induction,
p(n) is true ∀ n ∈ N

Question 11.
4ⁿ – 3n – 1 is divisible by 9.
Solution:
Let p(n) be the statement:
4ⁿ – 3n – 1 is divisible by 9
Since 4¹ – 3(1) – 1 = 0 is divisible by 9.
The statement is true for n = 1
Assume that p(n) is true for n = k
(i.e) 4^k – 3k – 1 is divisible by 9
Then 4^k – 3k – 1 = 9t, for some t ∈ N ……..(1)
Show that the statement p(n) is true for n = k + 1
(i.e.,) we show that S(k + 1) = 4^k+1 – 3(k+1) – 1 is divisible by 9
From (1), we have
4^k = 9t + 3k + 1
∴ S(k + 1) = 4 . 4^k – 3(k + 1) – 1
= 4(9t + 3k + 1) – 3k – 3 – 1
= 4(9t) + 9k
= 9[4t + k]
Hence s(k + 1) is divisible by 9
Since 4t + k is an integer
∴ 4^k+1 – 3(k+1) – 1 is divisible by 9
∴ The statement is true for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ k
(i.e.,) 4ⁿ – 3n – 1 is divisible by 9

Question 12.
3 . 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17.
Solution:
Let p(n) be the statement:
3. 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17
Since 3 . 5²⁽¹⁾⁺¹ + 2³⁽¹⁾⁺¹
= 3 . 5³ + 2⁴
= 3(125) + 16
= 375 + 16
= 391
= 17(23) is divisible by 17
∴ The statement is true for n = 1
Assume that the statement is true for n = k
(i.e) 3 . 5^2k+1 + 2^3k+1 is divisible by 17
Then 3 . 5^2k+1 + 2^3k+1 = 17t, for some t ∈ N ……..(1)
Show that the statement p(n) is true for n = k + 1
(i.e.,) We have to show that
\(\text { 3. } 5^{2(k+1)+1}+2^{3(k+1)+1}\) is divisible by 17
From (1) we have

Here 25t + 2^3k+1 is an integer
∴ \(\text { 3. } 5^{2(k+1)+1}+2^{3(k+1)+1}\) is divisible by 17
∴ The statement is true for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e.,) 3 . 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17.

Question 13.
1.2.3 + 2.3.4 + 3.4.5 + ……. upto n terms = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
Solution:
The nth term of the given series is (n) (n + 1) (n + 2)
Let p(n) be the statement:
1.2.3 + 2.3.4 + 3.4.5 +……. + (n) (n+1) (n+2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
and S(n) be the sum on the L.H.S.
∵ S(1) = 1.2.3 = \(\frac{(1)(1+1)(1+2)(1+3)}{4}\) = 1.2.3
∴ p(1) is true
Assume that the statement p(n) is true for n = k
(i.e) S(k) = 1.2.3 + 2.3.4 + 3.4.5 + ……. + k(k + 1) (k + 2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\)
We show that the statement is true for n = k + 1
(i.e) We show that S(k + 1) = \(\frac{(k+1)(k+2)(k+3)(k+4)}{4}\)
We observe that
S(k + 1) = 1.2.3 + 2.3.4 + …… + k(k + 1) (k + 2) + (k + 1) (k + 2) (k + 3)
= S(k) + (k + 1) (k + 2) (k + 3)
= \(\frac{k(k+1)(k+2)(k+3)}{4}\) + (k + 1)(k + 2)(k + 3)
= (k + 1)(k + 2)(k + 3) \(\left(\frac{k}{4}+1\right)\)
= \(\frac{(k+1)(k+2)(k+3)(k+4)}{4}\)
∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N
(i.e.,) 1.2.3 + 2.3.4 + 3.4.5 + ……. + (n)(n + 1)(n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)

Question 14.
\(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}\) + …. up to n terms = \(\frac{n}{24}\) [2n² + 9n + 13]
Solution:
The nth term of the given series is \(\frac{1^{3}+2^{3}+3^{3}+\ldots .+n^{3}}{1+3+5+\ldots+(2 n-1)}\)
Let p(n) be the statement :

and let S(n) be the sum on the L.H.S.
∵ S(1) = \(\frac{1^{3}}{1}=\frac{1}{24}(2+9+13)=1=\frac{1^{3}}{1}\)
∴ p(1) is true
Assume that p(k) is true
(i.e.,) S(k) = \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\ldots+\frac{1^{3}+2^{3}+\ldots \pm k^{3}}{1+3+\ldots+(2 k-1)}\) = \(\frac{k}{24}\) [2k² + 9k + 13]
We show that p(k + 1) is true
(i.e,) we show that


∴ The statement holds for n = k + 1
∴ By the principle of mathematical induction,
p(n) is true for all n
(i.e.,) \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\ldots+\frac{1^{3}+2^{3}+\ldots \ldots+n^{3}}{1+3+\ldots+(2 n-1)}\) = \(\frac{n}{24}\) [2n² + 9n + 13]

Question 15.
1² + (1² + 2²) + (1² + 2² + 3²) + ……. up to n terms = \(\frac{n(n+1)^{2}(n+2)}{12}\)
Solution:
The nth term of the given series is (1² + 2² + 3² + …… + n²)
Let p(n) be the statement:
1² + (1² + 2²) + (1² + 2² + 3²) + ……. + (1² + 2² + …… + n²) = \(\frac{\mathrm{n}(\mathrm{n}+1)^{2}(\mathrm{n}+2)}{12}\)
and the sum on the LH.S. is denoted by S(n).
Since S(1) = 1² = \(\frac{1(1+1)^{2}(1+2)}{12}\) = 1 = 1²
∴ p(1) is true.
Assume that the statement is true for n = k
(i.e.,) S(k) = 1² + (1² + 2²) + ……. + (1² + 2² + ……. + k²)
= \(\frac{k(k+1)^{2}(k+2)}{12}\)
We show that S(k + 1) = \(\frac{(k+1)(k+2)^{2}(k+3)}{12}\)
We observe that


∴ The statement holds for n = k + 1.
∴ By the principle of mathematical induction,
p(n) is true for all n ∈ N.
(i.e.,) 1² + (1² + 2²) + (1² + 2² + 3²) + …….. (1² + 2² + ………. + n²) = \(\frac{n(n+1)^{2}(n+2)}{12}\)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Inverse Trigonometric Functions Solutions Exercise 8(a)

I.

Question 1.
Evaluate the following.
(i) \(\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)\)
Solution:

(ii) \(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Solution:
\(\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\cos ^{-1}\left(\cos \frac{\pi}{4}\right)=\frac{\pi}{4}\)

(iii) sec^-1(-√2)
Solution:

(iv) cot^-1(-√3)
Solution:

(v) \(\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right)\)
Solution:

(vi) \(\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)\)
Solution:

(vii) \(\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)\)
Solution:

Question 2.
Find the values of
(i) \(\sin \left(\cos ^{-1} \frac{3}{5}\right)\)
Solution:
\(\sin \left(\cos ^{-1} \frac{3}{5}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)=\frac{4}{5}\)

(ii) \(\tan \left({cosec}^{-1} \frac{65}{63}\right)\)
Solution:
\(\tan \left({cosec}^{-1} \frac{65}{63}\right)=\tan \left(\tan ^{-1} \frac{63}{16}\right)\) = \(\frac{63}{16}\)

(iii) \(\sin \left(2 \sin ^{-1} \frac{4}{5}\right)\)
Solution:

(iv) \(\sin ^{-1}\left(\sin \frac{33 \pi}{7}\right)\)
Solution:

(v) \(\cos ^{-1}\left(\cos \frac{17 \pi}{6}\right)\)
Solution:

Question 3.
Simplify each of the following.
(i) \(\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]\)
Solution:

(ii) tan^-1(sec x + tan x)
Solution:

(iii) \(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)
Solution:

(iv) sin^-1(2 cos²θ – 1) + cos-1(1 – 2 sin²θ)
Solution:
sin^-1(cos 2θ) + cos^-1(cos 2θ)
= sin^-1[sin (90° – 2θ)] + cos^-1(cos 2θ)
= 90° – 2θ + 2θ
= 90°

(v) \(\tan ^{-1}\left(x+\sqrt{1+x^{2}}\right)\); x ∈ R
Solution:

II.

Question 1.
Prove that
(i) \(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{36}{85}\)
Solution:

(ii) \(\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}\)
Solution:

(iii) \(\tan \left[\cot ^{-1} 9+{cosec}^{-1} \frac{\sqrt{41}}{4}\right]=1\)
Solution:

(iv) \(\cos ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{\sqrt{34}}=\tan ^{-1} \frac{27}{11}\)
Solution:

Question 2.
Find the values of
(i) \(\sin \left(\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}\right)\)
Solution:

(ii) \(\tan \left(\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{5}{\sqrt{34}}\right)\)
Solution:

(iii) \(\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)\)
Solution:
Let \(\sin ^{-1} \frac{3}{5}\) = α and \(\sin ^{-1} \frac{5}{13}\) = β

Question 3.
Prove that
(i) \(\cos \left[2 \tan ^{-1} \frac{1}{7}\right]=\sin \left[2 \tan ^{-1} \frac{3}{4}\right]\)
Solution:

(ii) \(\tan \left[2 \tan ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\right]=2\)
Solution:

(iii) \(\cos \left\{2\left[\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)\right]\right\}=\frac{3}{5}\)
Solution:

Question 4.
Prove that
(i) \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}-\tan ^{-1} \frac{2}{9}=0\)
Solution:

(ii) \(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\)
Solution:

(iii) \(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}-\tan ^{-1} \frac{8}{19}=\frac{\pi}{4}\)
Solution:

(iv) \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}\) = \(\cot ^{-1} \frac{201}{43}+\cot ^{-1} 18\)
Solution:

Question 5.
Show that
(i) sec² (tan^-1 2) + cosec² (cot^-1 2) = 10
Solution:
Let a = tan^-1 2 ⇒ tan α = 2
sec² α = 1 + tan^-1 α = 1 + 4 = 5
Let β = cot^-1 2 ⇒ cot β = 2
cosec² β = 1 + cot² β = 1 + 4 = 5
LHS = sec² (tan^-1 2) + cosec² (cot^-1 2)
= 5 + 5
= 10
= RHS

(ii) Find the value of \(\left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)\)
Solution:

(iii) If sin^-1 x – cos^-1 x = \(\frac{\pi}{6}\) then find x.
Solution:

III.

Question 1.
Prove that
(i) \(2 \sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{5}{13}=\cos ^{-1} \frac{323}{325}\)
Solution:

(ii) \(\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{2}\)
Solution:

(iii) \(4 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{99}-\tan ^{-1} \frac{1}{70}=\frac{\pi}{4}\)
Solution:

Question 2.
(i) If α = \({tan}^{-1}\left(\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right)\) then prove that x² = sin 2α.
Solution:

(ii) Prove that tan\(\left\{2-\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\right\}=\mathbf{x}\)
Solution:

(iii) Prove that \(\sin \left[\cot ^{-1} \frac{2 x}{1-x^{2}}+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]\) = 1
Solution:

(iv) Prove that \(\left\{\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right\}\)
Solution:

Question 3.
(i) If cos^-1 p + cos^-1 q + cos^-1 r = π, then prove that p² + q² + r² + 2pqr = 1
Solution:
Let cos^-1 p = A, cos^-1 q = B and cos^-1 r = C
then A + B + C = π ………(1)
and p = cos A, q = cos B and r cos C
Now p² + q² + r² = cos² A + cos² B + cos² C
= cos² A + (1 – sin² B + cos² C)
= 1 + (cos² A – sin² B) + cos² C
= 1 + cos (A + B) . cos (A – B) + cos² C
= 1 + cos (π – C) cos (A – B) + cos² C (By (1))
= 1 – cos C cos (A – B) + cos² C
= 1 – cos C [cos (A – B) – cos C]
= 1 – cos C [cos (A – B) – cos(180° – \(\overline{A+B}\)]
= 1 – cos C [cos (A – B) + cos (A + B)]
= 1 – cos C [2 cos A cos B]
= 1 – 2 pqr
∴ p² + q² + r² + 2pqr = 1

(ii) If \(\sin ^{-1}\left(\frac{2 p}{1+p^{2}}\right)-\cos ^{-1}\left(\frac{1-q^{2}}{1+q^{2}}\right)\) = \(\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\), then prove that x = \(\frac{p-q}{1+p q}\)
Solution:

(iii) If a, b, c are distinct non-zero real numbers having the same sign, prove that \(\cot ^{-1}\left(\frac{a b+1}{a-b}\right)+\cot ^{-1}\left(\frac{b c+1}{b-c}\right)\) + \(\cot ^{-1}\left(\frac{c a+1}{c-a}\right)\) = π or 2π
Solution:
Since (a – b) + (b – c) + (c – a) = 0.
(a – b), (b – c), (c – a) all cannot have the same sign.
Now two cases arise, namely, either two of these numbers are positive and one negative (or) two of these numbers are negative and one is positive.
Case (i): Without loss of generality, we assume that (a – b), (b – c) are both positive and (c – a) is negative

Case (ii): Without loss of generality, we assume that (a – b) and (b – c) are both negative and (c – a) is positive.

(iv) If sin^-1 (x) + sin^-1 (y) + sin^-1 (z) = π, then prove that \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z\)
Solution:
Let sin^-1 (x) = A, sin^-1 (y) = B and sin^-1 (z) = C
Then A + B + C = π …………(1)
and x = sin A, y = sin B and z = sin C
Now LHS = \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}\)
= sin A \(\sqrt{1-\sin ^{2} A}\) + sin B \(\sqrt{1-\sin ^{2} B}\) + sin C \(\sqrt{1-\sin ^{2} C}\)
= sin A cos A + sin B cos B + sin C cos C
= \(\frac{1}{2}\) [sin 2A + sin 2B + sin 2C]
= \(\frac{1}{2}\) [2 . sin (A + B) cos (A – B) + sin 2C]
= \(\frac{1}{2}\) [2 sin (π – c). cos (A – B) + sin 2C]
= \(\frac{1}{2}\) [2 sin C cos (A – B) + 2 sin C cos C]
= \(\frac{1}{2}\) (2 sin C) [cos (A – B) + cos C]
= sin C [cos (A – B) + cos (180° – \(\overline{A+B}\)]
= sin C [cos (A – B) – cos (A + B)]
= sin C [2 sin A sin B]
= 2 xyz
∴ \(x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z\)

(v) (a) If tan^-1 x + tan^-1 y + tan^-1 z = π, then prove that x + y + z = xyz.
Solution:
Let A = tan^-1 x, B = tan^-1 y, C = tan^-1 z
tan A = x, tan B = y, tan C = z
Given A + B + C = π ……….(1)
A + B = π – C
tan (A + B) = tan (π – C)
\(\frac{\tan A+\tan B}{1-\tan A \tan B}\) = -tan C
\(\frac{x+y}{1-x y}\) = -z
x + y = -z + xyz
∴ x + y + z = xyz

(b) If tan^-1 x + tan^-1 y + tan^-1 z = \(\frac{\pi}{2}\), then prove that xy + yz + zx = 1.
Solution:
Let tan^-1 x = A, tan^-1 y = B and tan^-1 z = C
Then A + B + C = \(\frac{\pi}{2}\) …….(1)
and x = tan A, y = tan B and z = tan C
∵ A + B + C = \(\frac{\pi}{2}\)
A + B = \(\frac{\pi}{2}\) – C
⇒ tan (A + B) = tan(\(\frac{\pi}{2}\) – C)
⇒ \(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}\) = cot C
⇒ \(\frac{x+y}{1-x y}=\frac{1}{z}\)
⇒ zx + yz = 1 – xy (or) xy + yz + zx = 1

Question 4.
Solve the following equations for x:
(i) \({Tan}^{-1}\left(\frac{x-1}{x-2}\right)+{Tan}^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)
Solution:

(ii) \(\tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)\) = \(\tan ^{-1} \frac{2}{x^{2}}\)
Solution:
Given that

⇒ x²(3x + 1) = 2x(4x + 3)
⇒ x [x(3x + 1) – 2(4x + 3)] = 0
⇒ x = 0 (or) 3x² – 7x – 6 = 0
⇒ x = 0 (or) 3x² – 9x + 2x – 6 = 0
⇒ x = 0 (or) 3x(x – 3) + 2(x – 3) = 0
⇒ x = 0 (or) (3x + 2) (x – 3) = 0
⇒ x = 0 (or) 3 (or) \(\frac{-2}{3}\)

(iii) \(3 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\) + \(2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}\)
Solution:

(iv) sin^-1(1 – x) – 2 sin^-1 x = \(\frac{\pi}{2}\)
Solution:
Given sin^-1(1 – x) – 2 sin^-1x = \(\frac{\pi}{2}\)
Let sin^-1(1 – x) = α and sin^-1(x) = β
Then sin α = 1 – x and sin β = x
cos α = \(\sqrt{1-(1-x)^{2}}\) and cos β = \(\sqrt{1-x^{2}}\)
Now sin^-1(1 – x) – 2 sin^-1(x) = \(\frac{\pi}{2}\)
⇒ α – 2β = \(\frac{\pi}{2}\)
⇒ α = \(\frac{\pi}{2}\) + 2β
⇒ sin α = sin (\(\frac{\pi}{2}\) + 2β)
⇒ sin α = cos 2β
⇒ 1 – x = 1 – 2 sin2β
⇒ 1 – x = 1 – 2x²
⇒ 2x² – x = 0
⇒ x(2x – 1) = 0
⇒ x = 0 (or) x = \(\frac{1}{2}\)
But when x = \(\frac{1}{2}\)

Hence x = 0 is the only solution for the given equation.

Question 5.
Solve the following equations.
(i) \(\cot ^{-1}\left(\frac{1+x}{1-x}\right)=\frac{1}{2} \cot ^{-1}\left(\frac{1}{x}\right)\), x > 0 and x ≠ 1
Solution:

(ii) \(\tan \left[\cos ^{-1} \frac{1}{x}\right]=\sin \left[\cot ^{-1} \frac{1}{2}\right]\); x ≠ 0
Solution:
Let \(\cos ^{-1}\left(\frac{1}{x}\right)=\alpha, \cot ^{-1} \frac{1}{2}=\beta\)

(iii) \(\cos ^{-1} x+\sin ^{-1} \frac{x}{2}=\frac{\pi}{6}\)
Solution:

(iv) \(\cos ^{-1}(\sqrt{3} \cdot x)+\cos ^{-1} x=\frac{\pi}{2}\)
Solution:

(v) \(\sin \left[\sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x\right]=1\)
Solution:

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(d) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(d)

I.

Question 1.
Simplify
(i) \(\frac{\sin 2 \theta}{1+\cos 2 \theta}\)
Solution:
\(\frac{\sin 2 \theta}{1+\cos 2 \theta}\)
= \(\frac{2 \sin \theta \cos \theta}{2 \cos ^{2} \theta}\)
= tan θ

(ii) \(\frac{3 \cos \theta+\cos 3 \theta}{3 \sin \theta-\sin 3 \theta}\)
Solution:
\(\frac{3 \cos \theta+\cos 3 \theta}{3 \sin \theta-\sin 3 \theta}\)
= \(\frac{3 \cos \theta+4 \cos ^{3} \theta-3 \cos \theta}{3 \sin \theta-\left(3 \sin \theta-4 \sin ^{3} \theta\right)}\)
= \(\frac{4 \cos ^{3} \theta}{4 \sin ^{3} \theta}\)
= cot³θ

Question 2.
Evaluate the following.
(i) 6 sin 20° – 8 sin3 20°
Solution:
6 sin 20° – 8 sin³20°
= 2(3 sin 20° – 4 sin³20°)
= 2 sin (3 × 20)
= 2 sin 60°
= 2 \(\left[\frac{\sqrt{3}}{2}\right]\)
= √3

(ii) cos²72° – sin²54°
Solution:

(iii) sin²42° – sin²12°
Solution:
sin (42 + 12) sin (42 – 12)
= sin 54° . sin 30°
= \(\left[\frac{\sqrt{5}+1}{4}\right] \frac{1}{2}\)
= \(\frac{\sqrt{5}+1}{8}\)

Question 3.
(i) Express \(\frac{\sin 4 \theta}{\sin \theta}\) interms of cos³θ and cos θ.
Solution:
consider sin 4θ = sin(3θ + θ)
= sin 3θ cos θ + cos 3θ sin θ
= (3 sin θ – 4 sin³θ) cos θ + (4 cos³θ – 3 cos θ) sin θ
= 3 sin θ cos θ – 4 sin³θ cos θ + 4 cos³θ sin θ – 3 cos θ sin θ
= 4 cos³θ sin θ – 4 sin³θ cos θ
= sin θ (4 cos³θ – 4 sin²θ cos θ)
\(\frac{\sin 4 \theta}{\sin \theta}=\frac{\sin \theta\left(4 \cos ^{3} \theta-4 \sin ^{2} \theta \cos \theta\right)}{\sin \theta}\)
= 4 cos³θ – 4(1 – cos²θ) cos θ
= 4 cos³θ – 4 cos θ + 4 cos³θ
= 8 cos³θ – 4 cos θ

(ii) Express cos⁶A + sin⁶A in terms of sin 2A.
Solution:
cos⁶A + sin⁶A
= (cos²A)³ + (sin²A)³
= (cos²A + sin²A)³ – 3 cos²A sin²A (cos²A + sin²A)
= 1 – 3 cos²A sin²A
= 1 – \(\frac{3}{4}\) (4 cos²A sin²A)
= 1 – \(\frac{3}{4}\) (sin²2A)

(iii) Express \(\frac{1-\cos \theta+\sin \theta}{1+\cos \theta+\sin \theta}\) in terms of tan θ/2
Solution:

Question 4.
(i) If sin α = \(\frac{3}{5}\), where \(\frac{\pi}{2}\) < α < π, evaluate cos 3α and tan 2α.
Solution:

(ii) If cos A = \(\frac{7}{25}\) and \(\frac{3 \pi}{2}\) < A < 2π, then find the value of cot A/2.
Solution:

(iii) If 0 < θ < \(\frac{\pi}{8}\), show that \(\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 \theta)}}}=2 \cos (\theta / 2)\)
Solution:

Question 5.
Find the extreme values of
(i) cos 2x + cos2x
Solution:
cos 2x + cos²x
= 2 cos²x – 1 + cos²x
= 3 cos²x – 1
-1 ≤ cos x ≤ 1
0 ≤ cos²x ≤ 1
0 ≤ 3 cos²x ≤ 3
-1 ≤ 3 cos²x – 1 ≤ 2
maximum value = 2
minimum value = -1

(ii) 3 sin²x + 5 cos²x
Solution:
3 sin²x + 5 cos²x
= 3(1 – cos²x) + 5 cos²x
= 3 – 3 cos²x + 5 cos²x
= 3 + 2 cos²x
-1 ≤ cos x ≤ 1
0 ≤ cos²x ≤ 1
0 ≤ 2 cos²x ≤ 2
3 ≤ 3 + 2 cos²x ≤ 5
maximum value = 5
minimum value = 3

Question 6.
If a ≤ cos θ + 3√2 sin[θ + \(\frac{\pi}{4}\)] + 6 ≤ b, then find the largest value of a and smallest value of b.
Solution:
a ≤ cos θ + 3√2 sin[θ + \(\frac{\pi}{4}\)] + 6 ≤ b
consider cos θ + 3√2 sin[θ + \(\frac{\pi}{4}\)] + 6
= cos θ + 3√2[sin θ cos \(\frac{\pi}{4}\) + cos θ sin \(\frac{\pi}{4}\)] + 6
= cos θ + 3√2 sin θ \(\frac{1}{\sqrt{2}}\) + 3√2 cos θ \(\frac{1}{\sqrt{2}}\) + 6
= cos θ + 3 sin θ + 3 cos θ + 6
= 4 cos θ + 3 sin θ + 6
∴ a = 4, b = 3, c = 6
maximum value = c + \(\sqrt{a^{2}+b^{2}}\)
= 6 + \(\sqrt{16+9}\)
= 6 + 5
= 11
minimum value = c – \(\sqrt{a^{2}+b^{2}}\)
= 6 – 5
= 1

Question 7.
Find the periods for the following functions.
(i) cos⁴x
Solution:

(ii) \(2 \sin \left[\frac{\pi x}{4}\right]+3 \cos \left[\frac{\pi x}{3}\right]\)
Solution:

(iii) sin²x + 2 cos²x
Solution:
Let f(x) = sin²x + 2 cos²x
= 1 – cos²x + 2 cos²x
= 1 + cos²x
= 1 + \(\frac{1+\cos 2 x}{2}\)
∴ period of cos 2x = \(\frac{2 \pi}{2}\) = π
∴ period of f(x) = π

(iv) \(2 \sin \left[\frac{\pi}{4}+x\right] \cos x\)
Solution:

(v) \(\frac{5 \sin x+3 \cos x}{4 \sin 2 x+5 \cos x}\)
Solution:
Let f(x) = \(\frac{5 \sin x+3 \cos x}{4 \sin 2 x+5 \cos x}\)
period of sin x = 2π
period of cos x = 2π
period of sin 2x = \(\frac{2 \pi}{2}\) = π
period of cos x = 2π
L.C.M. of (2π, 2π, π, 2π) = 2π
∴ period of f(x) = 2π

II.

Question 1.
(i) If 0 < A < \(\frac{\pi}{4}\) and cos A = \(\frac{4}{5}\), find the values of sin 2A and cos 2A.
Solution:

(ii) For what values of A in the first quadrant, the expression \(\frac{\cot ^{3} A-3 \cot A}{3 \cot ^{2} A-1}\) is positive?
Solution:

(iii) Prove that \(\frac{\cos 3 A+\sin 3 A}{\cos A-\sin A}=1+2 \sin A\)
Solution:

Question 2.
(i) Prove that \(\cot \left[\frac{\pi}{4}-\theta\right]=\frac{\cos 2 \theta}{1-\sin 2 \theta}\) and hence find the value of cot 15°.
Solution:

(ii) If θ lies in third quadrant and sin θ = \(\frac{-4}{5}\), find the values of cosec (θ/2) and tan (θ/2).
Solution:

(iii) If 450° < θ < 540° and sin θ = \(\frac{12}{13}\), then calculate sin (θ/2) and cos (θ/2)
Solution:

(iv) Prove that \(\frac{1}{\cos 290^{\circ}}+\frac{1}{\sqrt{3} \sin 250^{\circ}}=\frac{4}{\sqrt{3}}\)
Solution:

Question 3.
Prove that
(i) \(\frac{\sin 2 A}{(1-\cos 2 A)} \cdot \frac{(1-\cos A)}{\cos A}=\tan \frac{A}{2}\)
Solution:

(ii) \(\frac{\sin 2 x}{(\sec x+1)} \cdot \frac{\sec 2 x}{(\sec 2 x+1)}\) = \(\tan \frac{x}{2}\)
Solution:

(iii) \(\frac{\left(\cos ^{3} \theta-\cos 3 \theta\right)}{\cos \theta}+\frac{\left(\sin ^{3} \theta+\sin 3 \theta\right)}{\sin \theta}=\mathbf{3}\)
Solution:

Question 4.
(i) Show that cos A = \(\frac{\cos 3 A}{(2 \cos 2 A-1)}\). Hence find the value of cos 15°.
Solution:

(ii) Show that sin A = \(\frac{\sin 3 A}{1+2 \cos 2 A}\). Hence find the value of sin 15°.
Solution:

(iii) Prove that tan α = \(\frac{\sin 2 \alpha}{1+\cos 2 \alpha}\) and hence deduce the values tan 15° and tan 22\(\frac{1^{\circ}}{2}\).
Solution:

Question 5.
Prove that
(i) \(\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}=4\)
Solution:

(ii) √3 cosec 20° – sec 20° = 4
Solution:

(iii) tan 9° – tan 27° – cot 27° + cot 9° = 4
Solution:

(iv) If \(\frac{\sin \alpha}{a}=\frac{\cos \alpha}{b}\), then prove that a sin 2α + b cos 2α = b
Solution:
Given \(\frac{\sin \alpha}{a}=\frac{\cos \alpha}{b}\)
b sin α = a cos α
L.H.S. = a sin 2α + b cos 2α
= a 2 sin α cos α + b (1 – 2 sin²α)
= 2 sin α (a cos α) + b – 2b sin²α
= 2 sin α (b sin α) + b – 2b sin²α
= 2b sin²α + b – 2b sin²α
= b

Question 6.
(i) In a ∆ABC; if \(\tan \frac{A}{2}=\frac{5}{6}\) and tan\(\tan \frac{B}{2}=\frac{20}{37}\), then show that \(\tan \frac{C}{2}=\frac{2}{5}\)
Solution:

(ii) If cos θ = \(\frac{5}{13}\) and 270° < θ < 360°, evaluate sin (θ/2) and cos (θ/2).
Solution:
cos θ = \(\frac{5}{13}\)
given 270 < θ < 360°
⇒ 135 < \(\frac{\theta}{2}\) < 180°
∴ θ lies in the fourth quadrant
θ/2 lies is second quadrant

(iii) If 180° < θ < 270° and sin θ = \(\frac{-4}{5}\) calculate sin[θ/2] cos[θ/2]
Solution:
Given sin θ = \(\frac{-4}{5}\)
⇒ cos θ = \(\frac{-3}{5}\)
given 180 < θ < 270
∴ θ in the III quadrant
Now 90 < \(\frac{\theta}{2}\) < 135
∴ \(\frac{\theta}{2}\) is in II quadrant

Question 7.
(i) Prove that \(\cos ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\cos ^{2} \frac{5 \pi}{8}+\cos ^{2} \frac{7 \pi}{8}\) = 2
Solution:

(ii) Show that \(\cos ^{4}\left(\frac{\pi}{8}\right)+\cos ^{4}\left(\frac{3 \pi}{8}\right)+\cos ^{4}\left(\frac{5 \pi}{8}\right)+\cos ^{4}\left(\frac{7 \pi}{8}\right)\) = \(\frac{3}{2}\)
Solution:

III.

Question 1.
(i) If tan x + tan(x + \(\frac{\pi}{3}\)) + tan(x + \(\frac{2 \pi}{3}\)) = 3, show that tan 3x = 1
Solution:

(ii) Prove that \(\sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \cdot \sin \frac{3 \pi}{5} \cdot \sin \frac{4 \pi}{5}\) = \(\frac{5}{16}\)
Solution:

(iii) Show that \(\cos ^{2}\left(\frac{\pi}{10}\right)+\cos ^{2}\left(\frac{2 \pi}{5}\right)+\cos ^{2}\left(\frac{3 \pi}{5}\right)\) + \(\cos ^{2}\left(\frac{9 \pi}{10}\right)\) = 2
Solution:

Question 2.
(i) \(\frac{1-\sec 8 \alpha}{1-\sec 4 \alpha}=\frac{\tan 8 \alpha}{\tan 2 \alpha}\)
Solution:

(ii) \(\left[1+\cos \frac{\pi}{10}\right]\left[1+\cos \frac{3 \pi}{10}\right]\left[1+\cos \frac{7 \pi}{10}\right]\) \(\left[1+\cos \frac{9 \pi}{10}\right]=\frac{1}{16}\)
Solution:

Question 3.
(i) Prove that \(\cos \frac{2 \pi}{7} \cdot \cos \frac{4 \pi}{7} \cdot \cos \frac{8 \pi}{7}=\frac{1}{8}\)
Solution:

(ii) \(\cos \frac{\pi}{11} \cdot \cos \frac{2 \pi}{11} \cdot \cos \frac{3 \pi}{11} \cdot \cos \frac{4 \pi}{11}\) \(\cos \frac{5 \pi}{11}=\frac{1}{32}\)
Solution:

Question 4.
(i) If cos α = \(\frac{3}{5}\) and cos β = \(\frac{5}{13}\) and α, β are acute angles, then prove that
(a) \(\sin ^{2}\left[\frac{\alpha-\beta}{2}\right]=\frac{1}{65}\)
(b) \(\cos ^{2}\left[\frac{\alpha+\beta}{2}\right]=\frac{16}{65}\)
Solution:

(ii) If A is not an integral multiple of π, prove that cos A . cos 2A . cos 4A . cos 8A = \(\frac{\sin 16 A}{16 \sin A}\) and hence deduce that \(\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}=\frac{1}{16}\)
Solution:
Take 16 sin A {cos A . cos 2A . cos 4A . cos 8A}
= 8(2 sin A . cos A) cos 2A . cos 4A . cos 8A
= 8 sin 2A . cos 2A . cos 4A . cos 8A
= 4(2 sin 2A . cos 2A) . cos 4A . cos 8A
= 4 sin 4A . cos 4A . cos 8A
= 2 (2 sin 4A . cos 4A) . cos 8A
= 2 sin 8A . cos 8A
= sin (16A)
∴ 16 sin A {cos A . cos 2A . cos 4A . cos 8A} = sin (16A)
∴ cos A . cos 2A . cos 4A . cos 8A = \(\frac{\sin (16 A)}{16 \sin A}\)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(b)

I. Find the periods for the given 1 – 5 functions.

Question 1.
cos(3x + 5) + 7
Solution:
f(x) = cos(3x + 5) + 7
We know that the function g(x) = cos x for all x ∈ R has the period 2π.
Now f(x) = cos(3x + 5) + 7
We get that f(x) is periodic and the period of f is \(\frac{2 \pi}{|3|}=\frac{2 \pi}{3}\)

Question 2.
tan 5x
Solution:
The function g(x) = tan x periodic and π is the period.
∴ f(x) = tan 5x periodic and its period is \(\frac{\pi}{|5|}=\frac{\pi}{5}\)

Question 3.
\(\cos \left(\frac{4 x+9}{5}\right)\)
Solution:
The function h(x) = cos x for all x ∈ R has the period 2π.
Now f(x) = \(\cos \left(\frac{4 x}{5}+\frac{9}{5}\right)\) is periodic and period of f is \(\frac{2 \pi}{\left(\frac{4}{5}\right)}=\frac{5 \pi}{2}\)

Question 4.
|sin x|
Solution:
The function h(x) = sin x for all x ∈ R has the period 2π.
But f(x) = |sin x| is periodic and its period is π.
∵ f(x + π) = |sin(x + π)|
= |-sin x|
= sin x

Question 5.
tan(x + 4x + 9x + …… + n²x) (n any positive integer)
Solution:
tan(1² + 2² + 3² + …… + n²) x = \(\tan \left[\frac{n(n+1)(2 n+1)}{6}\right] x\)
period = \(\frac{6 \pi}{n(n+1)(2 n+1)}\)

Question 6.
Find a sine function whose period is \(\frac{2}{3}\)
Solution:
\(\frac{2 \pi}{|k|}=\frac{2}{3}\)
3π = |k|
∴ sin kx = sin 3πx

Question 7.
Find a cosine function whose period is 7.
Solution:
\(\frac{2 \pi}{|k|}\) = 7
\(\frac{2 \pi}{7}\) = |k|
∴ cos kx = cos \(\frac{2 \pi}{7}\) x

II. Sketch the graph of the following functions.

Question 1.
tan x between 0 and \(\frac{\pi}{4}\)
Solution:

Question 2.
cos 2x in [0, π]
Solution:

Question 3.
sin 2x in the interval (0, π)
Solution:

Question 4.
sin x in the interval [-π, +π]
Solution:

Question 5.
cos²x in [0, π]
Solution:

III.

Question 1.
Sketch the region enclosed by y = sin x, y = cos x and X-axis in the interval [0, π].
Solution: