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

Question

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

Dr. Abubacker · Accepted Answer

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

Q. 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

Explanation

Reviewed by Dr. Abubacker

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