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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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Verified Answer
The correct answer is:
$-6$
$\cot \theta=\frac{-2}{3}$
$\therefore \quad \theta \in 2$ nd quadrant
$$
\begin{aligned}
& \therefore \sin \theta=\frac{3}{\sqrt{13}}, \cos \theta=\frac{-2}{\sqrt{13}}, \tan \theta=\frac{-3}{2} \\
& \therefore \quad \frac{(5 \sin \theta+\cos \theta)^2}{\tan \theta+\cot \theta}=\frac{\left(\frac{15}{\sqrt{13}}-\frac{2}{\sqrt{13}}\right)^2}{-\frac{2}{3}-\frac{3}{2}}=\frac{13}{\frac{-13}{6}}=-6 .
\end{aligned}
$$
$\therefore \quad \theta \in 2$ nd quadrant
$$
\begin{aligned}
& \therefore \sin \theta=\frac{3}{\sqrt{13}}, \cos \theta=\frac{-2}{\sqrt{13}}, \tan \theta=\frac{-3}{2} \\
& \therefore \quad \frac{(5 \sin \theta+\cos \theta)^2}{\tan \theta+\cot \theta}=\frac{\left(\frac{15}{\sqrt{13}}-\frac{2}{\sqrt{13}}\right)^2}{-\frac{2}{3}-\frac{3}{2}}=\frac{13}{\frac{-13}{6}}=-6 .
\end{aligned}
$$
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