${\cos }^{-1}\left(\frac{1}{2}\right) + 2{\sin }^{-1}\left(\frac{1}{2}\right)$ is equal to :

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{2\mathrm{\pi }}{3}$

A particle possess two velocities simultaneously at an angle of ${\tan }^{-1}\left(\frac{12}{5}\right)$; to each other. Their resultant is 15 m/s. If one velocity is 13 m/s, then the other will be :

• 5 m/s

• 4 m/s

• 12 m/s

• 13 m/s

243.

If ${\cos }^{-1}\left(\mathrm{p}\right) +\hspace{0.17em}{\cos }^{-1}\left(\mathrm{q}\right) +\hspace{0.17em}{\cos }^{-1}\left(\mathrm{r}\right) = \mathrm{\pi }$, then p² + q² + r² + 2pqr is equal to

• 3

• 1

• 2

• - 1

If ${\sin }^{-1}\left(\frac{\mathrm{x}}{5}\right) +\hspace{0.17em}{\csc }^{-1}\left(\frac{5}{4}\right) = \frac{\mathrm{\pi }}{2}$, then x is equal to

• 1

• 4

• 3

• 5

$\sin \left(\frac{1}{2}{\cos }^{-1}\left(\frac{4}{5}\right)\right)$

• $- \frac{1}{\sqrt{10}}$

• $\frac{1}{\sqrt{10}}$

• $- \frac{1}{10}$

• $\frac{1}{10}$

If ${\cos }^{-1}\left(\mathrm{x}\right) + {\cos }^{-1}\left(\mathrm{y}\right) + {\cos }^{-1}\left(\mathrm{z}\right) = 3\mathrm{\pi }$, then xy + yz + zx is equal to

• 1

• 0

• - 3

• 3

If ${\tan }^{-1}\left(\frac{\mathrm{x} + 1}{\mathrm{x} - 1}\right) + {\tan }^{-1}\left(\frac{\mathrm{x} - 1}{\mathrm{x}}\right) = {\tan }^{-1}\left(- 7\right)$, then the value of x is

• zero

• - 2

• 1

• 2

If ${\cos }^{-1}\left(\sqrt{\mathrm{p}}\right) + {\cos }^{-1}\left(\sqrt{1 - \mathrm{p}}\right) +\hspace{0.17em}{\cos }^{-1}\left(\sqrt{1 - \mathrm{q}}\right) = \frac{3\mathrm{\pi }}{4}$, then the value of q is

• $\frac{2}{\sqrt{2}}$

• 1

• 1/2

• 1/3

The soluton of ${\sin }^{-1}\left(\mathrm{x}\right) - {\sin }^{-1}\left(2\mathrm{x}\right) = \pm \frac{\mathrm{\pi }}{3}$ is

• $\pm \frac{1}{3}$

• $\pm \frac{1}{4}$

• $\pm \frac{\sqrt{3}}{2}$

• $\pm \frac{1}{2}$

If ${\cos }^{-1}\left(\mathrm{x}\right) = \mathrm{\alpha }, \left(0 < \mathrm{x} < 1\right)$ and ${\sin }^{-1}\left(2\mathrm{x}\sqrt{1 - {\mathrm{x}}^2}\right)$ + ${\sec }^{-1}\left(\frac{1}{2{\mathrm{x}}^2 - 1}\right) = \frac{2\mathrm{\pi }}{3}$, then ${\tan }^{-1}\left(2\mathrm{x}\right)$ equals :

• $\frac{\mathrm{\pi }}{6}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{2}$