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 $\cot \theta=-\sqrt{3} \ldots( i )$
Also, $\alpha$ lies in $3rd$ quadrant and $\sin \alpha=-\frac{3}{5}$
$\therefore \tan \alpha=\frac{3}{4}$ and $\cos \alpha=-\frac{4}{5} \ldots$ (ii)
$\therefore \frac{2 \tan \alpha+\sqrt{3} \tan \theta}{\cot ^{2} \theta+\cos \alpha}=\frac{2 \cdot \frac{3}{4}-\sqrt{3} \cdot \frac{1}{\sqrt{3}}}{3-\frac{4}{5}}$
$=\frac{\frac{3}{2}-1}{3-\frac{4}{5}}=\frac{5}{22}$