Question

If $\cos \theta =-\frac{\sqrt{3}}{2}$ and $\sin \alpha =-\frac{3}{5}$, where $\theta$ does not lie in the third quadrant, then $\frac{2\tan \alpha +\sqrt{3}\tan \theta }{{\cot }^2\theta +\cos \alpha }$ is equal to

• A. $\frac{7}{22}$
• B. $\frac{5}{22}$
• C. $\frac{9}{22}$
• D. $\frac{22}{5}$

Answer 1

Answer: (B)

Given, $\cos \theta =-\frac{\sqrt{3}}{2}<0$ and $\theta$ does not lie in third quadrant.
$\therefore \theta$ must be lying in $2nd$ quadrant.
$\Rightarrow \tan \theta =-\frac{1}{\sqrt{3}}$
and $\cot \theta =-\sqrt{3}\dots (i)$
Also, $\alpha$ lies in $3rd$ quadrant and $\sin \alpha =-\frac{3}{5}$
$\therefore \tan \alpha =\frac{3}{4}$ and $\cos \alpha =-\frac{4}{5}\dots$ (ii)
$\therefore \frac{2\tan \alpha +\sqrt{3}\tan \theta }{{\cot }^2\theta +\cos \alpha }=\frac{2\cdot \frac{3}{4}-\sqrt{3}\cdot \frac{1}{\sqrt{3}}}{3-\frac{4}{5}}$
$=\frac{\frac{3}{2}-1}{3-\frac{4}{5}}=\frac{5}{22}$