Question

$co{t}^{-1}\left(\sqrt{cos\alpha }\right)-ta{n}^{-1}\left(\sqrt{cos\alpha }\right)=x$ , then sin x is equal to

• A. ${\tan }^2\frac{\alpha }{2}$
• B. ${\cot }^2\frac{\alpha }{2}$
• C. $\tan \alpha$
• D. $\cot \frac{\alpha }{2}$

Answer 1

The correct option is A ${\tan }^2\frac{\alpha }{2}$

${\cot }^{-1}\left(\sqrt{\cos \alpha }\right)-{\tan }^{-1}\left(\sqrt{\cos \alpha }\right)=x$

$\frac{\pi }{2}-{\tan }^{-1}\left(\sqrt{\cos \alpha }\right)-{\tan }^{-1}\left(\sqrt{\cos \alpha }\right)=x$ (because ${\cot }^{-1}x+{\tan }^{-1}x=\frac{\pi }{2}$)

$\Rightarrow \frac{\pi }{2}-2{\tan }^{-1}\left(\sqrt{\cos \alpha }\right)=x$

$\Rightarrow \sin x=\sin (\frac{\pi }{2}-2{\tan }^{-1}(\sqrt{\cos \alpha }\left)\right)$

Let's say ${\tan }^{-1}\left(\sqrt{\cos \alpha }\right)=\theta$

$\Rightarrow \tan \theta =\sqrt{\cos \alpha }$

$\sin x=sin(\frac{\pi }{2}-2\theta )=\cos 2\theta$

$=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

$=\frac{1-\cos \alpha }{1+\cos \alpha }$

$\sin x=\frac{1-\cos \alpha }{1+\cos \alpha }$

$=\frac{2{\sin }^2\alpha /2}{2{\cos }^2\alpha /2}$

$\Rightarrow \sin x={\tan }^2\frac{\alpha }{2}$.