If the equation $2 \operatorname{Cot}^{-1}(x^2+2x+k) = \pi - 3 \operatorname{Tan}^{-1}(x^2+2x+k)$ has two distinct real solutions,then all the values of $k$ lie in the interval

If $f(x) = \cos \left( {{{\tan }^{ - 1}}\left( {\sin \left( {{{\cos }^{ - 1}}x} \right)} \right)} \right) + \sin \left( {{{\cot }^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}x} \right)} \right)} \right)$ has range $[m, M)$,then the number of solutions of the equation $\operatorname{sgn} (|x - 1| - 2) = \ln |x - 2|$ is (where $\operatorname{sgn}$ denotes the signum function).

The sum $\sum\limits_{n = 1}^\infty {{\cot }^{ - 1}} \left( {\frac{{2\left( {\sum\limits_{k = 1}^n k } \right) - 1}}{3}} \right)$ is equal to

If $\sin ^{-1} \frac{\alpha}{17}+\cos ^{-1} \frac{4}{5}-\tan ^{-1} \frac{77}{36}=0$ and $0 < \alpha < 13$,then $\sin ^{-1}(\sin \alpha)+\cos ^{-1}(\cos \alpha)$ is equal to $.........$.

The number of solutions of $\operatorname{Tan}^{-1} 1 + \frac{1}{2} \operatorname{Cos}^{-1} x^2 - \operatorname{Tan}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right) = 0$ is

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