The value of $\cos \left(45°\right)\cos \left(7\frac{1}{2}°\right)\sin \left(7\frac{1}{2}°\right)$ is 

• $\frac{1}{2}$

• $\frac{1}{8}$

• $\frac{1}{4}$

• $\frac{1}{16}$

General solution of $\sin \left(\mathrm{x}\right) + \cos \left(\mathrm{x}\right) = \underset{\mathrm{a}\in \mathrm{R}}{\min }\left\{1, {\mathrm{a}}^2 - 4\mathrm{a} + 6\right\}$ is

• $\frac{\mathrm{n\pi }}{2} + {\left(- 1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{4}$

• $2\mathrm{n\pi } + {\left(- 1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{4}$

• $\mathrm{n\pi } + {\left(- 1\right)}^{\mathrm{n} +\hspace{0.17em}1}\frac{\mathrm{\pi }}{4}$

• $\mathrm{n\pi } + {\left(- 1\right)}^{\mathrm{n}}\frac{\mathrm{\pi }}{4} - \frac{\mathrm{\pi }}{4}$

If a= 2$\sqrt{2}$, b = 6, A= 45°, then

• no triangle is possible

• one triangle is possible

• two triangles are possible

• either no triangle or two triangles are possible

In a triangle ABC, if $\sin \left(\mathrm{A}\right) \sin \left(\mathrm{B}\right) = \frac{\mathrm{ab}}{{\mathrm{c}}^2}$, then the triangle is

• equilateral

• isosceles

• right angled

• obtuse angled

The value of $\left(1 + \cos \left(\frac{\mathrm{\pi }}{6}\right)\right)\left(1 + \cos \left(\frac{\mathrm{\pi }}{3}\right)\right)\left(1 + \cos \left(\frac{2\mathrm{\pi }}{3}\right)\right)\left(1 + \cos \left(\frac{7\mathrm{\pi }}{6}\right)\right)$ is

• $\frac{3}{16}$

• $\frac{3}{8}$

• $\frac{3}{4}$

• $\frac{1}{2}$

If $\mathrm{P} = \frac{1}{2}{\sin }^2\left(\mathrm{\theta }\right) + \frac{1}{3}{\cos }^2\left(\mathrm{\theta }\right)$ then

• $\frac{1}{3} \le \mathrm{P} \le \frac{1}{2}$

• $\mathrm{P} \ge \frac{1}{2}$

• $2 \le \mathrm{P} \le 3$

• $- \frac{\sqrt{13}}{6} \le \mathrm{P} \le \frac{\sqrt{13}}{6}$

A positive acute angle is divided into two parts whose tangents are $\frac{1}{2}$ and $\frac{1}{3}$. Then, the angle is

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{5}$

• $\frac{\mathrm{\pi }}{3}$

• $\frac{\mathrm{\pi }}{6}$

The smallest value of $5\cos \left(\mathrm{\theta }\right) + 12$ is

• 5

• 12

• 7

• 17

Show that

$\frac{\sin \left(\mathrm{\theta }\right)}{\cos \left(3\mathrm{\theta }\right)} + \frac{{\displaystyle \sin \left(3\mathrm{\theta }\right)}}{{\displaystyle \cos \left(9\mathrm{\theta }\right)}} + \frac{{\displaystyle \sin \left(9\mathrm{\theta }\right)}}{{\displaystyle \cos \left(27\mathrm{\theta }\right)}} = \frac{1}{2}\left(\left(\tan 27\mathrm{\theta }\right) - \tan \left(\mathrm{\theta }\right)\right)$

90.

The equation $\sqrt{3}\sin \left(\mathrm{x}\right) + \cos \left(\mathrm{x}\right) = 4$ has

• infinitely many solutions

• no solution

• two solutions

• only one solution