If cos theta = -frac{sqrt{3}}{2} and sin alph | vedclass

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Question

(A) Given $\cos 	heta = -\frac{\sqrt{3}}{2}$. Since $	heta$ is not in the third quadrant and $\cos 	heta < 0$,$	heta$ must be in the second quadra

Vedclass · Accepted Answer

$\frac{5}{22}$

Vedclass · Answer

(A) Given $\cos 	heta = -\frac{\sqrt{3}}{2}$. Since $	heta$ is not in the third quadrant and $\cos 	heta In the second quadrant,$	an 	heta = -\frac{1}{\sqrt{3}}$.Given $\sin \alpha = -\frac{3}{5}$ and $\alpha$ is in the third quadrant $(180^{\circ} In the third quadrant,$	an \alpha = \frac{3}{4}$ and $\cos \alpha = -\frac{4}{5}$.Now,substitute these values into the expression $\frac{2 	an \alpha + \sqrt{3} 	an 	heta}{\cot^2 	heta + \cos \alpha}$:Numerator $= 2(\frac{3}{4}) + \sqrt{3}(-\frac{1}{\sqrt{3}}) = \frac{3}{2} - 1 = \frac{1}{2}$.Denominator $= (-\sqrt{3})^2 + (-\frac{4}{5}) = 3 - \frac{4}{5} = \frac{15-4}{5} = \frac{11}{5}$.Result $= \frac{1/2}{11/5} = \frac{1}{2} 	imes \frac{5}{11} = \frac{5}{22}$.

If $\cos \theta = -\frac{\sqrt{3}}{2}$ and $\sin \alpha = -\frac{3}{5}$,where $\theta$ does not lie in the third quadrant and $\alpha$ lies in the third quadrant,find the value of $\frac{2 \tan \alpha + \sqrt{3} \tan \theta}{\cot^2 \theta + \cos \alpha}$.

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