Given that the inverse trigonometric functions take principal values only. Then,the number of real values of $x$ which satisfy $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1} x$ is equal to:

If $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$, then $x \in$

Define $f: R \rightarrow R$ by $f(x) = \cos(\tan^{-1}(\sin(\tan^{-1} x)))$. Then $\lim_{x \rightarrow \infty} (f \circ f)(x)$ is equal to

$\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{(\cos ^{-1} x)^2} = $

For any $y \in R$,let $\cot ^{-1}(y) \in(0, \pi)$ and $\tan ^{-1}(y) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the sum of all the solutions of the equation $\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\cot ^{-1}\left(\frac{9-y^2}{6 y}\right)=\frac{2 \pi}{3}$ for $0 < |y| < 3$,is equal to

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 $...........$