Let $f(x) = \cot \left( \sin^{-1} \sqrt{\frac{2}{3 + \cos 2x}} \right)$. Then,the value of $f'\left( \frac{2\pi}{3} \right)$ is:

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):
$(A) \cos \beta > 0$
$(B) \sin \beta < 0$
$(C) \cos(\alpha + \beta) > 0$
$(D) \cos \alpha < 0$

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

The value of $\cos^{-1}(\cos 12) - \sin^{-1}(\sin 14)$ is

The number of solutions of $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$ is

Let $x = \sin(2 \tan^{-1} \alpha)$ and $y = \sin(\frac{1}{2} \tan^{-1} \frac{4}{3})$. If $S = \{\alpha \in R : y^2 = 1 - x\}$,then $\sum_{\alpha \in S} 16 \alpha^3$ is equal to $...........$