$\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x) =$ . . . . . .

Considering the principal values of the inverse trigonometric functions,$\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right)$,for $-\frac{1}{2} < x < \frac{1}{\sqrt{2}}$,is equal to:

Solve for $x$: $\tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2} \tan^{-1} x$,where $x > 0$.

$\frac{d}{dx} \left[ \tan^{-1} \left( \frac{a - x}{1 + ax} \right) \right] = $

Consider the following statements.
$I$. $\sin ^{-1}(y^2-4y+6)+\cos ^{-1}(y^2-4y+6) = \frac{\pi}{2}, \forall y \in R$
$II$. $\sec ^{-1}(y^2-4y+6)+\operatorname{cosec}^{-1}(y^2-4y+6) = \frac{\pi}{2}, \forall y \in R$
Which of the above statement$(s)$ is/are true?

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