Question

If $U={\cot }^{-1}\sqrt{\cos 2\theta }-{\tan }^{-1}\sqrt{\cos 2\theta }$, then $\sin U$ is equal to

• A. ${\sin }^2\theta$
• B. ${\cos }^2\theta$
• C. ${\tan }^2\theta$
• D. ${\tan }^{2}2\theta$

Answer 1

Answer: (C)

${\tan }^2\theta$

$U={\cot }^{-1}\sqrt{\cos 2\theta }-{\tan }^{-1}\sqrt{\cos 2\theta }$
$\Rightarrow U=\frac{\pi }{2}-{\tan }^{-1}\sqrt{\cos 2\theta }-{\tan }^{-1}\sqrt{\cos 2\theta }$
$\Rightarrow U=\frac{\pi }{2}-2{\tan }^{-1}\sqrt{\cos 2\theta }$
$\Rightarrow U=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{2\sqrt{\cos 2\theta }}{1-\cos 2\theta }\right)$
$\Rightarrow U=\frac{\pi }{2}-{\tan }^{-1}\left(\frac{\sqrt{\cos 2\theta }}{{\sin }^2\theta }\right)$
$\Rightarrow \tan (\frac{\pi }{2}-U)=\frac{\sqrt{\cos 2\theta }}{{\sin }^2\theta }$
$\Rightarrow \cot U=\frac{\sqrt{\cos 2\theta }}{{\sin }^2\theta }$
$\Rightarrow \sin U=\frac{1}{\sqrt{1+{\cot }^2\theta }}=\frac{1}{\sqrt{1+\frac{\cos \theta }{{\sin }^4\theta }}}$
$\Rightarrow \sin U=\frac{{\sin }^2\theta }{\sqrt{{\sin }^4\theta +1-2{\sin }^2\theta }}$
$\Rightarrow \sin U=\frac{{\sin }^2\theta }{|1-{\sin }^2\theta |}={\tan }^2\theta$