$$y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right), 0 \leq x < \frac{\pi}{2}$$, then $$\frac{d y}{d x}$$ at $$x=\frac{\pi}{6}$$ is

If $$y=\tan ^{-1}\left[\frac{1}{1+x+x^2}\right]+\tan ^{-1}\left[\frac{1}{x^2+3 x+3}\right], x>0$$, then $$\frac{d y}{d x}=$$

If $$\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1} \alpha$$, then $$\alpha=$$

$$\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$$, then the values of $$x$$ are