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If $tan^{-1}\frac{1-x}{1+x}=\frac{1}{2}tan^{-1}x$, then the value of x is |
$\frac{1}{2}$ $\frac{1}{\sqrt{3}}$ $\sqrt{3}$ 2 |
$\frac{1}{\sqrt{3}}$ |
We have $tan^{-1}\frac{1-x}{1+x}=\frac{1}{2}tan^{-1}x$ $⇒ tan^{-1}1-tan^{-1}x =\frac{1}{2}tan^{-1}x $ $⇒ \frac{\pi}{4}=\frac{3}{2}tan^{-1}x ⇒ tan^{-1}x =\frac{\pi}{6}⇒x = \frac{1}{\sqrt{3}}$ |
