# Solve for x : $tan^{-1} \bigg( \frac{2x}{1-x^2} \bigg) +cot^{-1} \bigg( \frac{1-x^2}{2x} \bigg) = \frac{\pi}{3}, x > 0$

Toolbox:
• Put $$x = tan\theta \Rightarrow \theta=tan^{-1}x$$
• $$\large\frac{2tan\theta}{1-tan^2\theta}=tan2\theta$$
• $$\large\frac{1-tan^2\theta}{2tan\theta}=cot2\theta$$
Put x=$$tan\theta$$ in L.H,S,
Then L.H.S.= $$\tan{-1}\large\frac{2tan\theta}{1-tan^2\theta}+cot^{-1}\large\frac{1-tan^2\theta}{2tan\theta}$$

Using the above formula $$\large\frac{2tan\theta}{1-tan^2\theta}=tan2\theta$$ and
$$\large\frac{1-tan^2\theta}{2tan\theta}=cot2\theta$$ we get
$$\Rightarrow tan^{-1}tan2\theta+cot^{-1}cot2\theta=\large\frac{\pi}{3}$$
$$\Rightarrow\:2\theta+2\theta=\large\frac{\pi}{3}$$
$$\Rightarrow 4\theta=\large\frac{\pi}{3}$$
$$\theta =\large \frac{\pi}{12}=tan^{-1}x$$
$$x = tan\large\frac{\pi}{12}$$

edited Mar 19, 2013