# PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

PSEB Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

## PSEB 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.

Evaluate:

(iii) cos 48° – sin 42°

= cos (90° – 42°) – sin 42°

[∵ cos (90° – 0) = sin O]

= sin 42° – sin 42° = 0.

(iv) cosec 31° – sec 59°

=cosec 31° – sec (90° – 31°)

= cosec 31° – cosec 31°

[∵ sec (90° – θ) = cosec θ].

Question 2.

Show that:

(i) tan 4 tan 230 tan 42° tan 67° = 1

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

Solution:

(ii) L.H.S.= cos 38° cos 52° – sin 38° sin 52°

= cos 38° × cos (90 – 38°) – sin 38° × sin (90° – 38°)

= cos 38° × sin 38° – sin 38° × cos 38

= 0.

∴ L.H.S. = RH.S.

Question 3.

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

Solution:

Given: tan 2A = cot (A – 18°)

⇒ cot (90° – 2A) = cot (A – 18°)

[cot (90° – θ) = tan θ]

⇒ 90°- 2A = A – 18°

⇒ 3A = 108°

⇒A = 36°.

Question 4.

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

Solution:

Given that: tan A = cot B

⇒ tan A = tan(90° – B)

[∵ tan (90° – θ) = cot θ]

⇒ A = 90° – B.

⇒ A + B = 90°..

Question 5.

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

Solution:

Given that: sec 4A = cosec (A – 20°)

⇒ cosec (90° – 4A) = cosec (A — 20°)

[∵ cosec (90° – θ) = sec θ]

⇒ 90° – 4A = A – 20°

⇒ 5A = 110°

⇒ A = 22°.

Question 7.

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

Solution:

Given that: sin 67° + cos 75°

= sin (90° – 23°) + cos (90° – 15°)

= cos 23° + sin 15°

[∵ sin(90° – θ) = cos θ and cos (90° – θ) = sin θ].

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