# Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry – Exercise 8.3

**Question 1. Evaluate :**

**(i) sin 18° / cos 72°**

**Solution:**

Since,

cos 72° = cos ( 90° – 18° ) = sin 18°

Therefore,

sin 18° / cos 72° = sin 18° / sin 18° = 1

Hence,

sin 18° / cos 72° = 1.

**(ii) tan 26° / cot 64°**

**Solution:**

Since,

cot 64° = cot ( 90° – 26° ) = tan 26°

Therefore,

tan 26° / cot 64° = tan 26° / tan 26° = 1

Hence,

tan 26° / cot 64° = 1.

**(iii) cos 48° – sin 42°**

Since,

cos 48° = cos ( 90° – 42° ) = sin 42°

Therefore,

cos 48° – sin 42° = sin 42° – sin 42° = 0

Hence,

cos 48° – sin 42° = 0.

**(iv) cosec 31° – sec 59°**

**Solution:**

Since,

sec 59° = sec ( 90° – 31° ) = cosec 31°

Therefore ,

cosec 31° – sec 59° = cosec 31° – cosec 31° = 0

Hence,

cosec 31° – sec 59° = 0.

**Question 2. Show that :**

**(i) tan 48° tan 23° tan 42° tan 67° = 1**

**Solution:**

Let A = tan 48° tan 23° tan 42° tan 67°

Since ,

tan 23° = tan( 90° – 23° ) = cot 67° and,

tan 42° = cot( 90° – 42° ) = cot 48°

Therefore,

A = tan 48° cot 67° cot 48° tan 67°

A = 1 (Since, tan B° cot B° = 1)

Hence,

tan 48° tan 23° tan 42° tan 67° = 1

** (ii) cos 38° cos 52° – sin 38° sin 52° = 0**

Let A = cos 38° cos 52° – sin 38° sin 52°

Since,

sin 52° = sin (90° – 38°) = cos 38° and,

cos 52° = cos(90° – 52°) = sin 38°

Therefore,

A = cos 38° sin 38° – sin 38° cos 38°

A = 0

Hence,

cos 38° cos 52° – sin 38° sin 52° = 0.

**Question 3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.**

**Solution:**

We have,

tan 2A = cot ( A – 18° ) —(1)

Since,

tan (2A) = cot ( 90° – 2A ) — (2)

Putting (2) in (1),

cot ( 90° – 2A ) = cot ( A – 18° )

Therefore,

90° – 2A = A – 18°

3A = 108°

A = 36°

Hence,

A = 36°.

**Question 4. If tan A = cot B, prove that A + B = 90°. **

**Solution:**

We have,

tan A = cot B —(1)

Since,

tan (A) = cot (90° – A) — (2)

Putting (2) in (1),

cot (90° – A) = cot (B)

Therefore,

90° – A = B

Hence,

A + B = 90°.

**Question 5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.**

**Solution:**

We have,

sec 4A = cosec ( A – 20° ) —(1)

Since,

sec 4A = cosec ( 90° – 4A ) — (2)

Putting (2) in (1),

cosec ( 90° – 4A ) = cosec ( A – 20° )

Therefore,

90° – 4A = A – 20°

5A = 110°

A = 22°

Hence,

A = 22°.

**Question 6. If A, B and C are interior angles of a triangle ABC, then show that sin ((B + C) / 2) = cos (A / 2).**

**Solution:**

Let T = sin ((B + C) / 2) — (1)

A, B and C are the interior angles of triangle ABC, therefore,

A + B + C = 180°

Dividing by 2 on both sides

(B + C)/2 = 90° – (A / 2) —(2)

Putting (2) on (1)

T = sin (90° – (A / 2)

= cos (A / 2)

Hence,

sin ((B + C)/2) = cos (A / 2).

**Question 7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°**

**Solution:**

Let A = sin 67° + cos 75°

Since,

sin 67° = sin(90° – 23°) = cos (23°)

cos 75° = cos (90° – 15°) = sin (15°)

Therefore,

sin 67° + cos 75° = cos 23° + sin 15°

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