સાબિત કરો કે $\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x$,જ્યાં $-\frac{1}{\sqrt{2}} \leq x \leq 1$.

(A) ધારો કે $x = \cos 2\theta$,તો $\theta = \frac{1}{2} \cos ^{-1} x$.
$L.H.S. = \tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$
$x = \cos 2\theta$ મૂકતા:
$= \tan ^{-1}\left(\frac{\sqrt{1+\cos 2\theta}-\sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta}+\sqrt{1-\cos 2\theta}}\right)$
નિત્યસમ $1+\cos 2\theta = 2\cos^2 \theta$ અને $1-\cos 2\theta = 2\sin^2 \theta$ નો ઉપયોગ કરતા:
$= \tan ^{-1}\left(\frac{\sqrt{2\cos^2 \theta}-\sqrt{2\sin^2 \theta}}{\sqrt{2\cos^2 \theta}+\sqrt{2\sin^2 \theta}}\right)$
$= \tan ^{-1}\left(\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta}\right)$
અંશ અને છેદને $\sqrt{2}\cos \theta$ વડે ભાગતા:
$= \tan ^{-1}\left(\frac{1 - \tan \theta}{1 + \tan \theta}\right)$
$\tan(\frac{\pi}{4} - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta}$ સૂત્રનો ઉપયોગ કરતા:
$= \tan ^{-1}\left(\tan\left(\frac{\pi}{4} - \theta\right)\right)$
$= \frac{\pi}{4} - \theta$
$\theta = \frac{1}{2} \cos ^{-1} x$ મૂકતા:
$= \frac{\pi}{4} - \frac{1}{2} \cos ^{-1} x = R.H.S.$