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Previous Year Question (PYQs)

If $\tan^{-1}\left(\dfrac{1-x}{1+x}\right)=\dfrac{1}{2}\tan^{-1}x$, then $x$ is




Solution

Let $t=\tan\left(\dfrac{1}{2}\tan^{-1}x\right)=\dfrac{1-x}{1+x}$. Using $\tan(2\theta)=\dfrac{2t}{1-t^2}=x$, we get $x=\dfrac{1-x^2}{2x}\Rightarrow 3x^2=1\Rightarrow x=\pm\dfrac{1}{\sqrt3}$. From the equation, $x>0$.

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