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}-\

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$\frac{\pi}{4}$

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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}-\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.

Similar Questions

જો $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$ હોય, તો $x$ ની કિંમત શોધો.

જો $\cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1-p} + \cos^{-1} \sqrt{1-q} = \frac{3\pi}{4}$ હોય,તો $q$ ની કિંમત શોધો.

$\cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)$ ની કિંમત શોધો.

શ્રેણીનો સરવાળો શોધો: $\tan^{-1} \left( \frac{1}{1 + 1 \times 2} \right) + \tan^{-1} \left( \frac{1}{1 + 2 \times 3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + n(n + 1)} \right) =$

$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)+\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)+\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ ની કિંમત શોધો.

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