# Prove that    11. $\tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right ) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x,\;\;-\frac{1}{\sqrt2}\leq x\leq 1$            [Hint: Put $x = \cos 2\theta$]

By using the Hint we will put $x = \cos 2\theta$;

we get then,

$=\tan^{-1}\left(\frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right )$

$=\tan^{-1}\left(\frac{\sqrt{2\cos^2 \theta} - \sqrt{2\sin^2\theta}}{\sqrt{2\cos^2 \theta} + \sqrt{2\sin^2\theta}} \right )$

$=\tan^{-1}\left(\frac{\sqrt2{\cos \theta} - \sqrt2{\sin\theta}}{\sqrt2{\cos \theta} + \sqrt2{\sin\theta}} \right )$

$=\tan^{-1}\left(\frac{{\cos \theta} - {\sin\theta}}{{\cos \theta} + {\sin\theta}} \right )$   dividing numerator and denominator by $\cos \theta$,

we get,

$= \tan^{-1}\left ( \frac{1-\tan \theta}{1+\tan \theta} \right )$

$= \tan^{-1} 1 - \tan^{-1} (\tan \theta)$      using the formula  $\left [ \tan^{-1}x - \tan^{-1} y = \tan^{-1} \frac{x-y}{1+xy} \right ]$

$= \frac{\pi}{4} - \theta = \frac{\pi}{4}- \frac{1}{2}\cos^{-1}x$

As L.H.S = R.H.S

Hence proved

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