Question

If $\cos \Theta =\frac{2}{3}$ and $\Theta$ is in the 4th quadrant , then $\frac{\sin \Theta +\tan \Theta }{\sin \Theta -\tan \Theta }$ is equal to

• A. $-\frac{1}{5}$
• B. $\frac{1}{5}$
• C. -5
• D. 5

Answer 1

The correct option is C -5

We have,

$\cos \theta =\frac{2}{3}$

Since, $\theta$ lies in $4th$ quadrant,

So,

$\sin \theta =-\frac{\sqrt{5}}{3}$

$\tan \theta =-\frac{\sqrt{5}}{2}$

Since,

$=\frac{\sin \theta +\tan \theta }{\sin \theta -\tan \theta }$

$=\frac{-\frac{\sqrt{5}}{3}-\frac{\sqrt{5}}{2}}{-\frac{\sqrt{5}}{3}+\frac{\sqrt{5}}{2}}$

$=\frac{\frac{1}{3}+\frac{1}{2}}{\frac{1}{3}-\frac{1}{2}}$

$=\frac{\frac{5}{6}}{\frac{-1}{6}}$

$=-5$

Hence, this is the answer.