Question

If $\theta$ = ${\cot }^{-1}\sqrt{\left(\cos x\right)}$ - ${\tan }^{-1}\sqrt{\left(\cos x\right)},$ then $\sin =\theta$

• A. $\tan \frac{x}{2}$
• B. ${\tan }^2\left(\frac{x}{2}\right)$
• C. $\frac{1}{2}{\tan }^{-1}\frac{x}{2}$
• D. None of these

Answer 1

Answer: (B)

$\theta ={\tan }^{-1}\left\{\frac{1}{\sqrt{\cos x}}\right\}-{\tan }^{-1}\sqrt{\cos x}$
$={\tan }^{-1}\left(\frac{\frac{1}{\sqrt{\cos x}}-\sqrt{\cos x}}{1+\left\{\frac{1}{\sqrt{\cos x}}.\sqrt{\cos x}\right\}}\right)$
or $\tan \theta =\frac{1-\cos x}{2\sqrt{\cos x}}$
$\therefore \tan \theta =\frac{1-\cos x}{2\sqrt{\cos x}}\therefore {\sin }^2\theta$
$=\frac{1}{1+{\cot }^2\theta }={\left(\frac{1-\cos x}{1+\cos x}\right)}^2={\tan }^4\left(\frac{x}{2}\right)$
$\Rightarrow \sin \theta ={\tan }^2\left(\frac{x}{2}\right)$