જો tan ^{-1}left(frac{1-x}{1+x}right)=frac{1} | vedclass

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(C) આપેલ સમીકરણ: $	an ^{-1}\left(\frac{1-x}{1+x}ight)=\frac{1}{2} 	an ^{-1} x$ નિત્યસમ $	an ^{-1} a - 	an ^{-1} b = 	an ^{-1}\left(\frac{a-b}{1

Vedclass · Accepted Answer

$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(C) આપેલ સમીકરણ: $	an ^{-1}\left(\frac{1-x}{1+x}ight)=\frac{1}{2} 	an ^{-1} x$ નિત્યસમ $	an ^{-1} a - 	an ^{-1} b = 	an ^{-1}\left(\frac{a-b}{1+ab}ight)$ નો ઉપયોગ કરતા,આપણે લખી શકીએ કે $	an ^{-1}(1) - 	an ^{-1}(x) = \frac{1}{2} 	an ^{-1} x$ આનું સાદું રૂપ આપતા $\frac{\pi}{4} = 	an ^{-1} x + \frac{1}{2} 	an ^{-1} x$ $\frac{\pi}{4} = \frac{3}{2} 	an ^{-1} x$ $	an ^{-1} x = \frac{\pi}{4} 	imes \frac{2}{3} = \frac{\pi}{6}$ તેથી,$x = 	an \left(\frac{\pi}{6}ight) = \frac{1}{\sqrt{3}}$

જો $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x$ હોય,તો $x$ ની કિંમત શોધો.

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