If $$2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$$, then the value of $$x$$ is

$$\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=$$

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