Prove that:
$$
\cos\left( \frac{3\pi}{2} + x \right)\cos(2\pi + x)\left[ \cot\left( \frac{3\pi}{2} - x \right) + \cot(2\pi + x) \right] = 1
$$

Q: Prove that: \[ \cos\left( \frac{3\pi}{2} + x \right)\cos(2\pi + x)\left[ \cot\left( \frac{3\pi}{2} - x \right) + \cot(2\pi + x) \right] = 1 \]

Explanation: \begin{align*} \text{LHS}&=\cos\left( \frac{3\pi}{2} + x \right)\cos(2\pi + x)\left[ \cot\left( \frac{3\pi}{2} - x \right) + \cot(2\pi + x) \right]\\ &= \sin x \cos x \left[ \tan x + \cot x \right] \\ &= \sin x \cos x \left( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right) \\ &= \sin x \cos x \cdot \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \\ &= \frac{\sin^2 x + \cos^2 x}{1} = 1 = \text{RHS} \qquad \textbf{Hence proved.} \end{align*}

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Accepted Answer

Explanation: 
\begin{align*}
	ext{LHS}&=\cos\left( \frac{3\pi}{2} + x ight)\cos(2\pi + x)\left[ \cot\left( \frac{3\pi}{2} - x ight) + \cot(2\pi + x) ight]\ &= \sin x \cos x \left[ 	an x + \cot x ight] \
&= \sin x \cos x \left( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} ight) \
&= \sin x \cos x \cdot \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \
&= \frac{\sin^2 x + \cos^2 x}{1} = 1 = 	ext{RHS}  \qquad 	extbf{Hence proved.}
\end{align*}

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