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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