Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a)

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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q1(i)

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

Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a)

(iii) sec-1(-√2)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q1(iii)

(iv) cot-1(-√3)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q1(iv)

(v) \(\sin \left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right)\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q1(v)

(vi) \(\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q1(vi)

(vii) \(\cos ^{-1}\left(\cos \frac{5 \pi}{4}\right)\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q1(vii)

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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q2(iii)

(iv) \(\sin ^{-1}\left(\sin \frac{33 \pi}{7}\right)\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q2(iv)

(v) \(\cos ^{-1}\left(\cos \frac{17 \pi}{6}\right)\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q2(v)

Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a)

Question 3.
Simplify each of the following.
(i) \(\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q3(i)

(ii) tan-1(sec x + tan x)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q3(ii)

(iii) \(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q3(iii)

(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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q3(v)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) I Q3(v).1

II.

Question 1.
Prove that
(i) \(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{8}{17}=\cos ^{-1} \frac{36}{85}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q1(i)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q1(i).1

(ii) \(\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q1(ii)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q1(ii).1

(iii) \(\tan \left[\cot ^{-1} 9+{cosec}^{-1} \frac{\sqrt{41}}{4}\right]=1\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q1(iii)

(iv) \(\cos ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{\sqrt{34}}=\tan ^{-1} \frac{27}{11}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q1(iv)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q1(iv).1

Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a)

Question 2.
Find the values of
(i) \(\sin \left(\cos ^{-1} \frac{3}{5}+\cos ^{-1} \frac{12}{13}\right)\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q2(i)

(ii) \(\tan \left(\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{5}{\sqrt{34}}\right)\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q2(ii)

(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
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q2(iii)

Question 3.
Prove that
(i) \(\cos \left[2 \tan ^{-1} \frac{1}{7}\right]=\sin \left[2 \tan ^{-1} \frac{3}{4}\right]\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q3(i)

(ii) \(\tan \left[2 \tan ^{-1}\left(\frac{\sqrt{5}-1}{2}\right)\right]=2\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q3(ii)

(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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q3(iii)

Question 4.
Prove that
(i) \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}-\tan ^{-1} \frac{2}{9}=0\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q4(i)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q4(i).1

(ii) \(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q4(ii)

(iii) \(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{3}{5}-\tan ^{-1} \frac{8}{19}=\frac{\pi}{4}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q4(iii)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q4(iii).1

(iv) \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}\) = \(\cot ^{-1} \frac{201}{43}+\cot ^{-1} 18\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q4(iv)

Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a)

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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q5(ii)

(iii) If sin-1 x – cos-1 x = \(\frac{\pi}{6}\) then find x.
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) II Q5(iii)

III.

Question 1.
Prove that
(i) \(2 \sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{5}{13}=\cos ^{-1} \frac{323}{325}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q1(i)

(ii) \(\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{2}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q1(ii)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q1(ii).1

(iii) \(4 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{99}-\tan ^{-1} \frac{1}{70}=\frac{\pi}{4}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q1(iii)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q1(iii).1

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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q2(i)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q2(i).1

(ii) Prove that tan\(\left\{2-\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\right\}=\mathbf{x}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q2(ii)

(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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q2(iii)

(iv) Prove that \(\left\{\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right\}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q2(iv)

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

Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a)

(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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q3(ii)

(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
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q3(iii)
Case (ii): Without loss of generality, we assume that (a – b) and (b – c) are both negative and (c – a) is positive.
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q3(iii).1

(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

Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a)

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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q4(i)

(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
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q4(ii)
⇒ 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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q4(iii)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q4(iii).1

(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 = \(\frac{1}{2}\)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q4(iv)
Hence x = 0 is the only solution for the given equation.

Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a)

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:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q5(i)

(ii) \(\tan \left[\cos ^{-1} \frac{1}{x}\right]=\sin \left[\cot ^{-1} \frac{1}{2}\right]\); x ≠ 0
Solution:
Let \(\cos ^{-1}\left(\frac{1}{x}\right)=\alpha, \cot ^{-1} \frac{1}{2}=\beta\)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q5(ii)

(iii) \(\cos ^{-1} x+\sin ^{-1} \frac{x}{2}=\frac{\pi}{6}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q5(iii)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q5(iii).1

(iv) \(\cos ^{-1}(\sqrt{3} \cdot x)+\cos ^{-1} x=\frac{\pi}{2}\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q5(iv)
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q5(iv).1

(v) \(\sin \left[\sin ^{-1}\left(\frac{1}{5}\right)+\cos ^{-1} x\right]=1\)
Solution:
Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions Ex 8(a) III Q5(v)