tan ^{-1}left{frac{sqrt{1+x}-sqrt{1-x}}{sqrt{ | vedclass

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(B) ધારો કે $x = \cos 2	heta$. તેથી $	heta = \frac{1}{2} \cos^{-1} x$. પદમાં $x = \cos 2	heta$ મૂકતા: $	an^{-1}\left(\frac{\sqrt{1+\cos 2	heta}-\

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) ધારો કે $x = \cos 2	heta$. તેથી $	heta = \frac{1}{2} \cos^{-1} x$. પદમાં $x = \cos 2	heta$ મૂકતા: $	an^{-1}\left(\frac{\sqrt{1+\cos 2	heta}-\sqrt{1-\cos 2	heta}}{\sqrt{1+\cos 2	heta}+\sqrt{1-\cos 2	heta}}ight)$ $= 	an^{-1}\left(\frac{\sqrt{2\cos^2	heta}-\sqrt{2\sin^2	heta}}{\sqrt{2\cos^2	heta}+\sqrt{2\sin^2	heta}}ight)$ $= 	an^{-1}\left(\frac{\sqrt{2}\cos	heta - \sqrt{2}\sin	heta}{\sqrt{2}\cos	heta + \sqrt{2}\sin	heta}ight)$ $= 	an^{-1}\left(\frac{1 - 	an	heta}{1 + 	an	heta}ight)$ $= 	an^{-1}(	an(\frac{\pi}{4} - 	heta))$ $= \frac{\pi}{4} - 	heta = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$. હવે,બીજું પદ ઉમેરતા: $(\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x) + \frac{1}{2} \cos^{-1} x = \frac{\pi}{4}$.

$\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}+\frac{1}{2} \cos ^{-1} x$ ની કિંમત શોધો.

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