$\tan \left(\cos ^{-1} \frac{1}{\sqrt{2}}+\tan ^{-1} \frac{1}{2}\right) = $

Considering only the principal values of the inverse trigonometric functions,the set $\{x \geq 0 : \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\}$

If $y = \tan ^{-1}\left(\frac{1}{1+x+x^{2}}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+2x+3}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+5x+7}\right) + \dots + n \text{ terms}$,then $y'(0)$ is

If $y = \sum_{k=1}^{6} k \cos^{-1} \left\{ \frac{3}{5} \cos kx - \frac{4}{5} \sin kx \right\}$,then $\frac{dy}{dx}$ at $x = 0$ is

If $\sum_{k=1}^n \tan^{-1} \left( \frac{1}{k^2+k+1} \right) = \tan^{-1} ( \theta )$,then $\theta =$

Find $\frac{dx}{dy}$ if $y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)$ for $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$.