Question

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

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

Answer 1

Answer: (A)

Put $\sqrt{cos\alpha }=\theta$
$\therefore co{t}^{-1}\theta -ta{n}^{-1}\theta =x$
$\Rightarrow \frac{\pi }{2}-2ta{n}^{-1}\theta =x$
$\therefore sinx=sin\left(\frac{\pi }{2}-2ta{n}^{-1}\theta \right)$
$=cos\left[2ta{n}^{-1}\theta \right]=cos\left[2t\right]$
where $ta{n}^{-1}\theta =t$
$\Rightarrow \theta =tant$
$=\frac{1-ta{n}^{2}t}{1+ta{n}^{2}t}=\frac{1-cos\alpha }{1+cos\alpha }$
$=\frac{2si{n}^2\frac{\alpha }{2}}{2co{s}^2\frac{\alpha }{2}}$
$=ta{n}^2\frac{\alpha }{2}$.