If $\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4},$ then find the value of $x$.

If $\tan^{-1} \left[ \frac{\sqrt{5} - 2\sqrt{6}}{1 + \sqrt{6}} \right] = \frac{\pi}{3} - \tan^{-1}(k)$, then $\sec^{-1}(k) = \dots$

If $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$ and $4x^2 = \frac{a}{b}$,then $a + b =$

If $3 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)-4 \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\frac{\pi}{3}$,then the value of $x=$

If $x_1, x_2, x_3$ are the roots of the equation $x^3 - x^2 \tan \theta + x \tan^2 \theta + \tan \theta = 0$ and $0 < \theta < \frac{\pi}{4}$, then the value of $\tan^{-1} x_1 + \tan^{-1} x_2 + \tan^{-1} x_3$ at $\theta = \frac{\pi}{12}$ is

Find $\frac{dy}{dx}$ for $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ where $|x| < 1$.