If $y = \tan^{-1}\left(\frac{1}{1+x+x^2}\right) + \tan^{-1}\left(\frac{1}{x^2+3x+3}\right) + \tan^{-1}\left(\frac{1}{x^2+5x+7}\right)$,then the value of $y'(0)$ is

Solve $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \csc x)$

If $\alpha = 3 \sin^{-1}\left(\frac{6}{11}\right)$ and $\beta = 3 \cos^{-1}\left(\frac{4}{9}\right)$,where the inverse trigonometric functions take only the principal values,then the correct option$(s)$ is(are):
$(A) \cos \beta > 0$
$(B) \sin \beta < 0$
$(C) \cos(\alpha + \beta) > 0$
$(D) \cos \alpha < 0$

If $x, y, z$ are in Arithmetic Progression and $\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are also in Arithmetic Progression,where $x, z > 0$ and $xz < 1, y < 1$,then

Find $\frac{dy}{dx}$ if $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $0 < x < 1$.

The number of real solutions of $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} 2x = \frac{\pi}{4}$ is