The value of $\sin \left[2{\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)\right]$ is

• $\frac{\sqrt{5}}{3}$

• $\frac{2\sqrt{5}}{3}$

• $\frac{4\sqrt{5}}{9}$

• $\frac{2\sqrt{5}}{9}$

$\sin \left(2{\sin }^{-1}\left(\sqrt{\frac{63}{65}}\right)\right)$ is equal to

• $\frac{2\sqrt{126}}{65}$

• $\frac{4\sqrt{65}}{65}$

• $\frac{8\sqrt{63}}{65}$

• $\frac{\sqrt{63}}{65}$

${\cot }^{-1}\left(2 . 1^2\right) + {\cot }^{-1}\left(2 . 2^2\right) + {\cot }^{-1}\left(2 . 3^2\right) + ...$ upto $\infty$ is equal to

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{5}$

If 'x' takes negative permissible value, then sin^-1(x) is equal to

• $- {\cos }^{-1}\left(\sqrt{1 - {\mathrm{x}}^2}\right)$

• ${\cos }^{-1}\left(\sqrt{{\mathrm{x}}^2 - 1}\right)$

   

• $\mathrm{\pi } - {\cos }^{-1}\left(\sqrt{1 - {\mathrm{x}}^2}\right)$

• ${\cos }^{-1}\left(\sqrt{1 - {\mathrm{x}}^2}\right)$

If a > b > 0, ${\sec }^{-1}\left(\frac{\mathrm{a} + \mathrm{b}}{\mathrm{a} - \mathrm{b}}\right) = 2{\sin }^{-1}\left(\mathrm{x}\right)$, then x is

• $- \sqrt{\frac{\mathrm{b}}{\mathrm{a} +\hspace{0.17em}\mathrm{b}}}$

• $\sqrt{\frac{\mathrm{b}}{\mathrm{a} +\hspace{0.17em}\mathrm{b}}}$

• $- \sqrt{\frac{\mathrm{a}}{\mathrm{a} +\hspace{0.17em}\mathrm{b}}}$

• $\sqrt{\frac{\mathrm{a}}{\mathrm{a} +\hspace{0.17em}\mathrm{b}}}$

If $\mathrm{x} \ne \mathrm{n\pi }, \mathrm{x} \ne \left(2\mathrm{n} + 1\right)\frac{\mathrm{\pi }}{2}, \mathrm{n} \in \mathrm{Z}$, then $\frac{{\sin }^{-1}\left(\cos \left(\mathrm{x}\right)\right) + {\cos }^{-1}\left(\sin \left(\mathrm{x}\right)\right)}{{\tan }^{-1}\left(\cot \left(\mathrm{x}\right)\right) + {\cot }^{-1}\left(\tan \left(\mathrm{x}\right)\right)}$ is

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{3}$

127.

In $∆\mathrm{ABC}$, if a = 2, B = ${\tan }^{-1}\left(\frac{1}{2}\right)$ and C = ${\tan }^{-1}\left(\frac{1}{3}\right)$, then (A, b) equals

• $\frac{3\mathrm{\pi }}{4}, \frac{2}{\sqrt{5}}$

• $\frac{\mathrm{\pi }}{4}, \frac{2\sqrt{2}}{\sqrt{5}}$

• $\frac{3\mathrm{\pi }}{4}, \frac{2\sqrt{2}}{\sqrt{5}}$

• $\frac{\mathrm{\pi }}{4}, \frac{2}{\sqrt{5}}$

The domain of f(x) = ${\sin }^{-1}\left[{\log }_2\left(\frac{\mathrm{x}}{2}\right)\right]$ is

• $0 \le \mathrm{x} \le 1$

• $0 \le \mathrm{x} \le 4$

• $1 \le \mathrm{x} \le 4$

• $4 \le \mathrm{x} \le 6$

If ${\tan }^{-1}\left(\mathrm{x}\right) = \frac{\mathrm{\pi }}{4} - {\tan }^{-1}\left(\frac{1}{3}\right)$, then x is

• $\frac{1}{3}$

• $\frac{1}{2}$

• $\frac{1}{4}$

• $\frac{1}{6}$

$\mathrm{If} {\cos }^{-1}\left(\frac{\mathrm{y}}{\mathrm{b}}\right) = \mathrm{nlog}\left(\frac{\mathrm{x}}{\mathrm{n}}\right)$, then

• ${\mathrm{xy}}_1 = \mathrm{n}\sqrt{{\mathrm{b}}^2 - {\mathrm{y}}^2}$

• ${\mathrm{xy}}_1 + \mathrm{n}\sqrt{{\mathrm{b}}^2 - {\mathrm{y}}^2} = 0$

• ${\mathrm{y}}_1 = \mathrm{x}\sqrt{{\mathrm{b}}^2 - {\mathrm{y}}^2}$

• ${\mathrm{xy}}_1 - \sqrt{{\mathrm{b}}^2 - {\mathrm{y}}^2} = 0$