Prove that:
$$
\frac{\cos(\pi + x)\cos(-x)}{\sin(\pi - x)\cos\left( \frac{\pi}{2} + x \right)} = \cot^2 x
$$

Q: Prove that: \[ \frac{\cos(\pi + x)\cos(-x)}{\sin(\pi - x)\cos\left( \frac{\pi}{2} + x \right)} = \cot^2 x \]

Explanation: \begin{align*} \text{LHS} &=\frac{\cos(\pi + x)\cos(-x)}{\sin(\pi - x)\cos\left( \frac{\pi}{2} + x \right)}\\&= \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} \\ &= \frac{-\cos^2 x}{-\sin^2 x} \\ &= \frac{\cos^2 x}{\sin^2 x} = \cot^2 x = \text{RHS} \qquad \textbf{Hence proved.} \end{align*}

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Accepted Answer

Explanation: 
\begin{align*}
	ext{LHS} &=\frac{\cos(\pi + x)\cos(-x)}{\sin(\pi - x)\cos\left( \frac{\pi}{2} + x ight)}\&= \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} \
&= \frac{-\cos^2 x}{-\sin^2 x} \
&= \frac{\cos^2 x}{\sin^2 x} = \cot^2 x = 	ext{RHS} \qquad 	extbf{Hence proved.}
\end{align*}

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