inverse trigonometric functions

The doamin of ${\sin }^{-1}\left[{\log }_2\left(\frac{\mathrm{x}}{12}\right)\right]$ is

• [2, 12]

• [- 1, 1]

• $\left[\frac{1}{3}, 24\right]$

• [6, 24]

If ${\tan }^{-1}\left(\mathrm{x} + 2\right) + {\tan }^{-1}\left(\mathrm{x} - 2\right) - {\tan }^{-1}\left(\frac{1}{2}\right) = 0$, then one of the values of x is equal to

• - 1

• 5

• $\frac{1}{2}$

• 1

The value of ${\sin }^{-1}\left(\frac{4}{5}\right) + 2{\tan }^{-1}\left(\frac{1}{3}\right)$ is equal to

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{6}$

The value of ${\sin }^{-1}\left(\cos \left(4095°\right)\right)$ is equal to

• $- \frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{6}$

• $- \frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{4}$

If $2{\sin }^{-1}\left(\mathrm{x}\right) - {\cos }^{-1}\left(\mathrm{x}\right) = \frac{\mathrm{\pi }}{2}$, then x is equal to

• $\frac{1}{\sqrt{2}}$

• $- \frac{1}{\sqrt{2}}$

• $\frac{- \sqrt{3}}{2}$

• $\frac{\sqrt{3}}{2}$

The value of ${\tan }^{-1}\left(2\right) + {\tan }^{-1}\left(3\right)$ is equal to

• $\frac{3\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{3}$

• ${\tan }^{-1}\left(6\right)$

The value of ${\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right) + {\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ is equal to

• ${\tan }^{-1}\left(\frac{5}{\sqrt{3}}\right)$

• ${\tan }^{-1}\left(\frac{2}{\sqrt{3}}\right)$

• ${\tan }^{-1}\left(\frac{1}{2}\right)$

• ${\tan }^{-1}\left(\frac{1}{3\sqrt{3}}\right)$

The value of $\cos \left({\sin }^{-1}\left(\frac{2}{3}\right)\right)$ is equal to

• $\frac{\sqrt{3}}{5}$

• $\frac{5}{3}$

• $\frac{5}{\sqrt{3}}$

• $\frac{\sqrt{5}}{3}$

If ${\tan }^{-1}\left(2\mathrm{x}\right) + {\tan }^{-1}\left(3\mathrm{x}\right) = \frac{\mathrm{\pi }}{2}$, then the value of x is equal to

• $\frac{1}{\sqrt{6}}$

• $\frac{1}{6}$

• $\frac{1}{\sqrt{3}}$

• $\frac{1}{\sqrt{2}}$

If ${\cos }^{-1}\left(\frac{1 - {\mathrm{x}}^2}{1 + {\mathrm{x}}^2}\right) + {\cos }^{-1}\left(\frac{1 - {\mathrm{y}}^2}{1 + {\mathrm{y}}^2}\right) = \frac{\mathrm{\pi }}{2}$, where xy < 1, then

• x - y - xy = 1

• x - y + xy = 1

• x + y - xy = 1

• x + y + xy = 1