NCERT Solutions class 10 maths chapter 8

Introduction to Trigonometry

NCERT Solutions class 10 maths chapter 8 Exercise 8.3

Question 1

Evaluate :

(i) sin 18°/cos 72°

(ii) tan 26°/cot 64°

(iii) cos 48° – sin 42°

(iv) cosec 31° – sec 59°

Answer

(i) sin 18°/cos 72°

We know that,sin (90° – A)= cos A.

= sin (90° – 18°) /cos 72°

= cos 72° /cos 72° = 1

(ii) tan 26°/cot 64°

We know that tan (90° – A) = cot A

= tan (90° – 36°)/cot 64°

= cot 64°/cot 64° = 1

(iii) cos 48° – sin 42°

We know that cos (90° – A)= sin A.

= cos (90° – 42°) – sin 42°

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

We know that,cosec (90° – 59°)= sec A

= cosec (90° – 59°) – sec 59°

= sec 59° – sec 59° = 0

Question 2

Show that:

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

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

Answer

(i) tan 48° tan 23° tan 42° tan 67°

We know that, tan 48° = tan (90° – 42°) = cot 42° , tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°

We know that,

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90°-38°) = sin 38°

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

= sin 52° sin 38° – sin 38° sin 52° = 0

Question 3

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

Answer

tan 2A = cot (A- 18°)

We know that tan 2A = cot (90° – 2A)

⇒ cot (90° – 2A) = cot (A -18°)

On compare

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

the value of A = 36°

Question 4

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

Answer

tan A = cot B

We know that cot B = tan (90° – B)

tan A = tan (90° – B)

A = 90° – B

A + B = 90°

Hence Proved.

Question 5

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

Answer

sec 4A = cosec (A – 20°)

We know that sec 4A = cosec (90° – 4A)

cosec (90° – 4A) = cosec (A – 20°)

Now, On compare

90° – 4A= A- 20°

110° = 5A

A = 110°/ 5 = 22°

Therefore, the value of 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:

We know that sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

⇒ B + C = 180° – A

On dividing both side by 2

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

Now, multiply both sides by sin functions, we get

⇒ sin (B+C)/2 = sin (90°-A/2)

Since sin (90°-A/2) = = cos A/2

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

Hence proved.

Question 7

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

Answer

Given:

sin 67° + cos 75°

sin 67° = sin (90° – 23°)

cos 75° = cos (90° – 15°)

= sin (90° – 23°) + cos (90° – 15°)

= cos 23° + sin 15°

Therefore, sin 67° + cos 75° is also expressed as cos 23° + sin 15°

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