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ID:
Class: 11
Subject: Math
Topic: Trigonometric Functions
Type: Short
Question:
Prove that: \[ \frac{\cos(\pi + x)\cos(-x)}{\sin(\pi - x)\cos\left( \frac{\pi}{2} + x \right)} = \cot^2 x \]
Official Solution
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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