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

### (i) sin 18Â° / cos 72Â°

Solution:

Since,

cos 72Â°  = cos ( 90Â° – 18Â° ) = sin 18Â°

Therefore,

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

Hence, sin 18Â° / cos 72Â°  = 1.

### (ii) tan 26Â° / cot 64Â°

Solution:

Since,

cot 64Â°  = cot ( 90Â° – 26Â° ) = tan 26Â°

Therefore,

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

Hence, tan 26Â° / cot 64Â°  = 1.

### (iii) cos 48Â° – sin 42Â°

Since,

cos 48Â°  = cos ( 90Â° – 42Â° ) = sin 42Â°

Therefore,

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

Hence, cos 48Â° – sin 42Â°  = 0.

### (iv) cosec 31Â° – sec 59Â°

Solution:

Since,

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

Therefore ,

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

Hence, cosec 31Â° – sec 59Â°  = 0.

### (i) tan 48Â° tan 23Â° tan 42Â° tan 67Â° = 1

Solution:

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

Since ,

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

tan  42Â° = cot(  90Â° –  42Â° ) = cot  48Â°

Therefore,

A = tan 48Â° cot 67Â° cot  48Â° tan  67Â°

A = 1  (Since, tan BÂ° cot  BÂ° = 1)

Hence, tan 48Â° tan 23Â° tan 42Â° tan 67Â° = 1

### (ii) cos 38Â° cos 52Â° â€“ sin 38Â° sin 52Â° = 0

Let A = cos 38Â° cos 52Â° â€“ sin 38Â° sin 52Â°

Since,

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

cos  52Â° = cos(90Â° –  52Â°) = sin 38Â°

Therefore,

A = cos 38Â° sin 38Â° â€“ sin 38Â° cos 38Â°

A = 0

Hence, cos 38Â° cos 52Â° â€“ sin 38Â° sin 52Â° = 0.

### Question 3. If tan 2A = cot (A â€“ 18Â°), where 2A is an acute angle, find the value of A.

Solution:

We have,

tan 2A = cot ( A – 18Â° )  —(1)

Since,

tan (2A) = cot ( 90Â° – 2A )  — (2)

Putting (2) in (1),

cot ( 90Â° – 2A ) = cot ( A – 18Â° )

Therefore,

90Â° – 2A = A – 18Â°

3A = 108Â°

A = 36Â°

Hence, A = 36Â°.

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

Solution:

We have,

tan A = cot B —(1)

Since,

tan (A) = cot (90Â° – A)  — (2)

Putting (2) in (1),

cot (90Â° – A) = cot (B)

Therefore,

90Â° – A = B

Hence, A + B = 90Â°.

### Question 5. If sec 4A = cosec (A â€“ 20Â°), where 4A is an acute angle, find the value of A.

Solution:

We have,

sec 4A = cosec ( A – 20Â° )  —(1)

Since,

sec 4A = cosec ( 90Â° – 4A )  — (2)

Putting (2) in (1),

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

Therefore,

90Â° – 4A = A – 20Â°

5A = 110Â°

A = 22Â°

Hence, A = 22Â°.

### 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 = 180Â°

Dividing by 2 on both sides

(B + C)/2  = 90Â° – (A / 2) —(2)

Putting (2) on (1)

T = sin (90Â° – (A / 2)

= cos (A / 2)

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

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

Solution:

Let A = sin 67Â° + cos 75Â°

Since,

sin 67Â° = sin(90Â° – 23Â°) = cos (23Â°)

cos 75Â° = cos (90Â° – 15Â°) = sin (15Â°)

Therefore,

sin 67Â° + cos 75Â° = cos 23Â° + sin 15Â°

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