Considering only the principal values of the inverse trigonometric functions,the set $\{x \geq 0 : \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\}$

For $a>0$, if $f(x)=ax+b$ is an onto function from $[-1,1]$ to $[0,2]$, then $\cot \left[\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}+\tan ^{-1} \frac{1}{5}\right]=$

If $y = \tan ^{-1}\left(\frac{1}{1+x+x^{2}}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+2x+3}\right) + \tan ^{-1}\left(\frac{1}{x^{2}+5x+7}\right) + \dots + n \text{ terms}$,then $y'(0)$ is

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).

Identify the pair$(s)$ of functions which are identical.

The number of real solutions of the equation $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2}$ is