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Question
Prove that: \[\cot^2 x - \tan^2 x = 4 \cot 2 x \text{ cosec } 2 x\]
Solution
\[LHS = \cot^2 x - \tan^2 x\]
\[ = \frac{\cos^2 x}{\sin^2 x} - \frac{\sin^2 x}{\cos^2 x}\]
\[ = \frac{\left( \cos^2 x \right)^2 - \left( \sin^2 x \right)^2}{\sin^2 x \cos^2 x}\]
\[= \frac{\left( \cos^2 x + \sin^2 x \right)\left( \cos^2 x - \sin^2 x \right)}{\sin^2 x \cos^2 x}\]
\[ = \frac{1 \times \left( \cos2x \right)}{\sin^2 x \cos^2 x} \left( \because \cos^2 x - \sin^2 x = \cos2x \right)\]
\[ = \frac{4\cos2x}{4 \sin^2 x \cos^2 x}\]
\[= \frac{4\left( \cos2x \right)}{\left( \sin2x \right)^2}\]
\[ = \frac{4\left( \cos2x \right)}{\left( \sin2x \right)} \times \frac{1}{\left( \sin2x \right)}\]
\[ = 4\cot2x \text{ cosec } 2x = RHS\]
\[\text{ Hence proved } .\]
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