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$\cot^{-1}\big(\sqrt{\cos\alpha}-\tan^{-1}(\sqrt{\cos\alpha}\big)=x$ then $\sin x=?$

$\begin{array}{1 1}(a)\;\tan^2\big(\large\frac{\alpha}{2}\big)&(b)\;\cot^2\big(\large\frac{\alpha}{2}\big)\\(c)\;\tan\alpha&(d)\;\cot\big(\large\frac{\alpha}{2}\big)\end{array}$

1 Answer

$\cot^{-1}\sqrt{\cos x}=\tan^{-1}\big(\sqrt{\cos\alpha}\big)=x$
$\tan^{-1}\big(\large\frac{1}{\sqrt{\cos\alpha}}\big)$$=\tan^{-1}(\sqrt{\cos\alpha})=x$
$\Rightarrow \tan^{-1}\bigg(\large\frac{\Large\frac{1}{\sqrt \cos\alpha}-\sqrt{\cos\alpha}}{1+\Large\frac{1}{\sqrt{\cos\alpha}}.\sqrt{\cos\alpha}}\bigg)=$$x$
$\Rightarrow \tan^{-1}\large\frac{1-\cos\alpha}{2\sqrt{\cos \alpha}}$$=x$
$\tan x=\large\frac{1-\cos x}{2\sqrt{\cos\alpha}}$
$\cot x=\large\frac{2\sqrt{\cos\alpha}}{1-\cos\alpha}$
$\sin x=\large\frac{1-\cos\alpha}{1+\cos\alpha}$
$\qquad=\large\frac{1-(1-2\sin^2\Large\frac{\alpha}{2})}{1+2\cos^2\Large\frac{\alpha}{2}-1}$
$\qquad=\large\frac{2\sin^2\Large\frac{\alpha}{2}}{2\cos^2\Large\frac{\alpha}{2}}$
(Or) $\sin x=\tan^2\large\frac{\alpha}{2}$
hence (a) is the correct answer.
answered Oct 16, 2013 by sreemathi.v
 

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