If cot theta = -frac{2}{3} and theta does not | vedclass

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(B) Given $\cot 	heta = -\frac{2}{3}$.Since $\cot 	heta < 0$ and $	heta$ is not in the $4^{	ext{th}}$ quadrant,$	heta$ must lie in the $2^{	ext{

Vedclass · Accepted Answer

$-6$

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(B) Given $\cot 	heta = -\frac{2}{3}$.Since $\cot 	heta In the $2^{	ext{nd}}$ quadrant,$\sin 	heta > 0$ and $\cos 	heta Using $\cot 	heta = \frac{\cos 	heta}{\sin 	heta} = -\frac{2}{3}$,let $\cos 	heta = -2k$ and $\sin 	heta = 3k$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $(3k)^2 + (-2k)^2 = 1 \implies 13k^2 = 1 \implies k = \frac{1}{\sqrt{13}}$.Thus,$\sin 	heta = \frac{3}{\sqrt{13}}$,$\cos 	heta = -\frac{2}{\sqrt{13}}$,and $	an 	heta = -\frac{3}{2}$.Substituting these values into the expression:$\frac{(5 \sin 	heta + \cos 	heta)^2}{	an 	heta + \cot 	heta} = \frac{(\frac{15}{\sqrt{13}} - \frac{2}{\sqrt{13}})^2}{-\frac{3}{2} - \frac{2}{3}} = \frac{(\frac{13}{\sqrt{13}})^2}{-\frac{9+4}{6}} = \frac{13}{-\frac{13}{6}} = -6$.

If $\cot \theta = -\frac{2}{3}$ and $\theta$ does not lie in the $4^{\text{th}}$ quadrant,then $\frac{(5 \sin \theta + \cos \theta)^2}{\tan \theta + \cot \theta} = $

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