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 cot^{-1}\theta - tan^{-1}\theta = x $
$\Rightarrow \frac{\pi}{2}-2\,tan^{-1}\theta = x $
$\therefore sin \,x = sin \left(\frac{\pi}{2} - 2\,tan^{-1}\theta\right) $
$= cos\left[ 2\,tan^{-1}\theta\right] = cos\left[2t\right] $
where $ tan^{-1}\theta = t $
$\Rightarrow \theta = tan \,t $
$= \frac{1-tan^{2} t }{1+tan^{2} t } =\frac{ 1-cos\, \alpha}{1+cos \,\alpha}$
$ =\frac{ 2 \,sin^{2} \frac{\alpha}{2}}{2\, cos^{2} \frac{\alpha}{2}}$
$ = tan^{2} \frac{\alpha}{2}$.