# AP SSC 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.3

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.3 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.3

### 10th Class Maths 11th Lesson Trigonometry Ex 11.3 Textbook Questions and Answers

Question 1.
Evaluate:
i) $$\frac{\tan 36^{\circ}}{\cot 54^{\circ}}$$
Given that $$\frac{\tan 36^{\circ}}{\cot 54^{\circ}}$$
= $$\frac{\tan 36^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}$$ [∵ cot (90 – θ) = tan θ]
= $$\frac{\tan 36^{\circ}}{\tan 36^{\circ}}$$
= 1

ii) cos 12° – sin 78°
Given that cos 12° – sin 78°
= cos 12° – sin(90° – 12°) [∵ sin (90 – θ) = cos θ]
= cos 12° – cos 12° = 0

iii) cosec 31° – sec 59°
Given that cosec 31° – sec 59°
= cosec 31° – sec (90° – 31°) [∵ sec (90 – θ) = cosec θ]
= cosec 31° – cosec 31° = 0

iv) sin 15° sec 75°
Given that sin 15° sec 75°
= sin 15° . sec (90° – 15°)
= sin 15° . cosec 15° [∵ sec (90 – θ) = cosec θ]
= sin 15° . $$\frac{1}{\sin 15^{\circ}}$$ [∵ cosec θ = $$\frac{1}{\sin \theta}$$]
= 1

v) tan 26° tan 64°
Given that tan 26° tan 64°
= tan 26° . tan (90° – 26°)
= tan 26° . cot 26° [∵ tan (90 – θ) = cot θ]
= tan 26° . $$\frac{1}{\tan 26^{\circ}}$$ [∵ cot θ = $$\frac{1}{\tan \theta}$$]
= 1

Question 2.
Show that
i) tan 48° tan 16° tan 42° tan 74° = 1
L.H.S. = tan 48° tan 16° tan 42° tan 74°
= tan 48°. tan 16° . tan(90° – 48°) . tan(90° – 16°)
= tan 48° . tan 16° . cot 48° . cot 16° [∵ tan (90 – θ) = cot θ]
= tan 48° . tan 16° . $$\frac{1}{\tan 48^{\circ}}$$ . $$\frac{1}{\tan 16^{\circ}}$$ [∵ cot θ = $$\frac{1}{\tan \theta}$$]
= 1 = R.H.S.
∴ L.H.S. = R.H.S.

ii) cos 36° cos 54° – sin 36° sin 54° = 0
L.H.S. = cos 36° cos 54° – sin 36° sin54°
= cos (90° – 54°) . cos (90° – 36°) – sin 36° . sin 54° [∵ cos (90 – θ) = sin θ]
= sin 54° . sin 36° – sin 36° . sin 54°
= 0 = R.H.S.
∴ L.H.S. = R.H.S.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle. Find the value of A.
Given that tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵ tan θ = cot (90 – θ)]
⇒ 90° – 2A = A – 18°
⇒ 108° = 3A
⇒ A = $$\frac{108^{\circ}}{3}$$ = 36°
Hence the value of A is 36°.

Question 4.
If tan A = cot B where A and B. are acute angles, prove that A + B = 90°.
Given that tan A = cot B
⇒ cot (90° – A) = cot B [∵ tan θ = cot (90 – θ)]
⇒ 90° – A = B
⇒ A + B = 90°

Question 5.
If A, B and C are interior angles of a triangle ABC, then show that $$\tan \left(\frac{\mathbf{A}+\mathbf{B}}{2}\right)=\cot \frac{\mathbf{C}}{2}$$