Maharashtra Board Class 11 Maths Part 1 Chapter 3 Trigonometry – II Ex 3.3 Solution

Question 1.
Find the values of:

Question 2.

Question 3.

v. tan x + cot x = 2 cosec 2x
Solution:
L.H.S. = tan x + cot x

Solution:

Solution:

= 2 cos x
= R.H.S.
[Note : The question has been modified.]

viii. 16 sin θ cos θ cos 2θ cos 4θ cos 8θ = sin 16θ
Solution:
L.H.S. = 16 sin θ cos θ cos 2θ cos 4θ cos 8θ
= 8(2sinθ cosθ) cos2θ cos 4θ cos 8θ
= 8sin 2θ cos 2θ cos 4θ cos 8θ
= 4(2sin 2θ cos 2θ) cos 4θ cos 8θ
= 4sin 4θ cos 4θ cos 8θ
= 2(2sin 4θ cos 4θ) cos 8θ
= 2sin 8θ cos 8θ
= sin 16θ
= R.H.S.

ix. \( = 2 cot 2x
Solution:

x. [latex]\frac{\cos x}{1+\sin x}=\frac{\cot \left(\frac{x}{2}\right)-1}{\cot \left(\frac{x}{2}\right)+1}\)
Solution:

Solution:

Solution:

Solution:

Solution:

xvii. 2 cosec 2x + cosec x = sec cot x/2
Solution:

Solution: