# Gujarat Board Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Gujarat Board Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.

Question 3.

If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A. (CBSE 2012)

Solution:

We have

tan 2A cot (A – 18°)

tan 2A = tan [90°- (A – 18°)]

tan 2A = tan [90° – A + 18°]

2A = 90° – A + 18°

3A = 108°

A = 108°/3

A = 36°

Question 4.

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

Solution:

We have

tan A = cot B

tanA = tan(90° – B) [cot θ = tan (90° – 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:

We have

sec 4A = cosec (A – 20°)

cosec (90° – 4A) = cosec (A – 20°)

90° – 4A = A – 20°

90° + 20° = 5A

5A = 110°

A = 110°/5 = 22°

Question 7.

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

Solution:

We have

sin 67° + cos 75°

= sin (90° – 23°) + cos (90° – 15°)

= cos 23° + sin 15°

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