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

  • Last Updated : 25 Jan, 2021

Question 1. Evaluate :

(i) sin 18° / cos 72°

Solution:

Since, 

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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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