Question

If ${\cot }^{-1}\left[{\left(\cos \alpha \right)}^{1/2}\right]+\left[{\tan }^{-1}{\left(\cos \alpha \right)}^{1/2}\right]=x$ , then $\sin x$ equals

• A. $1$
• B. ${\cot }^2\left(\frac{\alpha }{2}\right)$
• C. $\tan \alpha$
• D. $\cot \left(\frac{\alpha }{2}\right)$

Answer 1

The correct option is A $1$

As ${\tan }^{-1}\theta +{\cot }^{-1}\theta =\frac{\pi }{2}$
$\therefore x=\frac{\pi }{2}$
Hence,
$\sin x=\sin \frac{\pi }{2}=1$