If ² x x+² x- 2x =1+ x+ x, x (0, π), then x

Correct Answer is : 7π12,9π127π/12, 9π/12127π,129π

Correct Answer is : 7π12,9π127π/12, 9π/12127π,129π

Test Index

UPSC NDA Math Model Paper 2

Question : 56 of 120

Marks: +1, -0

\sin^2 x \tan x+\cos^2 \cot x-\sin 2x

=1+\tan x+\cot x, x \in (0, \pi),

x

Solution:

Given:

\sin^2 x \tan x+\cos^2 \cot x-\sin 2x

=1+\tan x+\cot x

\Rightarrow \frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}-\sin 2x

=1+\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}

\Rightarrow \frac{\sin^4 x+\cos^4 x}{\sin x \cos x}-\sin 2x

=1+\frac{\sin^2 x+\cos^2 x}{\sin x \cos x}

\Rightarrow \frac{1-2\sin^2 x \cos^2 x}{\sin x \cos x}-\sin 2x

=1+\frac{1}{\sin x \cos x}

\Rightarrow \frac{1}{\sin x \cos x}-\frac{2\sin^2 x \cos^2 x}{\sin x \cos x}-\sin 2x

\Rightarrow 2x=\frac{7\pi}{6}, \frac{11\pi}{6}

⇒ - 2 sin 2x = 1

⇒ sin 2x = - 1/2

\Rightarrow x=\frac{7\pi}{12}, \frac{11\pi}{12}

\frac{3\pi}{12}, \frac{5\pi}{12}

Hence, option C is the correct answer.

Go to Question:

Prev Question

Next Question