NIMCET 2016 — Mathematics PYQ

Concept:
$$\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$$
$$\cos \theta =\frac{\text{base}}{\text{hypotenuse}}$$
$$\tan \theta =\frac{\text{perpendicular}}{\text{base}}$$
Hypotenuse${}^2={\text{Perpendicular}}^2+{\text{Base}}^2$$

Calculation:

Given
\cos \theta = \frac{5}{13}, \frac{3\pi}{2} < \theta < 2\pi

From above $\theta$ lies in the fourth quadrant
so the $\tan \theta$ will be negative

Hypotenuse = 13
Base = 5

Perpendicular = $\sqrt{13^2+5^2}=\pm 12$

$\tan \theta =\frac{\text{perpendicular}}{\text{base}}$

$\tan \theta =\frac{-12}{5}$, $\tan \theta$ is negative in the fourth quadrant

$\tan 2\theta =\frac{2\times -12}{5}$

$\tan 2\theta =\frac{120}{119}$