If r² = x² + y² + z², then prove that

If cos(A) + cos(B) + cos(C) = 0, prove that
cos(3A) + cos(3B) + cos(3C) = 12 cos(A) cos(B)cos(C)

If $\sin \left(\mathrm{\theta }\right) = \frac{2\mathrm{t}}{1 + {\mathrm{t}}^2}$ and $\mathrm{\theta }$ lies in the second quadrant, then $\cos \left(\mathrm{\theta }\right)$ is equal to 

• $\frac{1 - {\mathrm{t}}^2}{1 + {\mathrm{t}}^2}$

• $\frac{{\mathrm{t}}^2 - 1}{1 + {\mathrm{t}}^2}$

• $\frac{- \left|1 - {\mathrm{t}}^2\right|}{1 + {\mathrm{t}}^2}$

• $\frac{1 + {\mathrm{t}}^2}{\left|1 - {\mathrm{t}}^2\right|}$

54.

The solutions set of in equation ${\cos }^{-1}\left(\mathrm{x}\right) < {\sin }^{-1}\left(\mathrm{x}\right)$

• [- 1, 1]

• $\left[\frac{1}{\sqrt{2}}, 1\right]$

• [0, 1]

• $(\frac{1}{\sqrt{2}}, 1]$

The number of solutions of $2\sin \left(\mathrm{x}\right) + \cos \left(\mathrm{x}\right) = 3$

• 1

• 2

• infinite

• no solution

Let $\tan \left(\mathrm{\alpha }\right) = \frac{\mathrm{a}}{\mathrm{a} + 1} \mathrm{and} \tan \left(\mathrm{\beta }\right) = \frac{1}{2\mathrm{a} + 1}$, then $\mathrm{\alpha } + \mathrm{\beta }$ is

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{3\mathrm{\pi }}{4}$

If $\mathrm{\theta } + \mathrm{\varphi } = \frac{\mathrm{\pi }}{4}, \mathrm{then} \left(1 + \tan \left(\mathrm{\theta }\right)\right) \left(1 + \tan \left(\mathrm{\varphi }\right)\right)$ is equal to

• 1

• 2

• 5/2

• 1/3

If $\sin \left(\mathrm{\theta }\right) \mathrm{and} \cos \left(\mathrm{\theta }\right)$ are the roots of the equation ax² - bx + c = 0, then a, b and c satisfy the relation

• a² + b² + 2ac = 0

• a² - b² + 2ac = 0

• a² + c² + 2ab = 0

• a² - b² - 2ac = 0

If $\sin \left(\mathrm{\theta }\right) + \cos \left(\mathrm{\theta }\right) = 0 \mathrm{and} 0 < \mathrm{\theta } < \mathrm{\pi }, \mathrm{then} \mathrm{\theta }$

• 0

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{3\mathrm{\pi }}{4}$

The value of $\cos \left(15°\right) - \sin \left(15°\right)$ is

• 0

• $\frac{1}{\sqrt{2}}$

• $- \frac{1}{\sqrt{2}}$

• $\frac{1}{2\sqrt{2}}$