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प्रश्न
Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]
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उत्तर
\[LHS = \tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right)\]
\[ = \frac{\tan\frac{\pi}{4} + \text{ tan } x}{1 - \tan\frac{\pi}{4}\text{ tan } x} + \frac{\tan\frac{\pi}{4} - \text{ tan } x}{1 + \tan\frac{\pi}{4}\text{ tan } x} \left[ \because \tan\left( A + B \right) = \frac{\text{ tan } A + \text{ tan } B}{1 - \text{ tan } A\text{ tan } B} \text{ and } \tan\left( A - B \right) = \frac{\text{ tan }A - \text{ tan } B}{1 + \text{ tan } A\text{ tan } B} \right]\]
\[= \frac{1 + \text{ tan } x}{1 - \text{ tan } x} + \frac{1 - \text{ tan } x}{1 + \text{ tan } x}\]
\[ = \frac{\left( 1 + \text{ tan } x \right)^2 + \left( 1 - \text{ tan } x \right)^2}{\left( 1 + \text{ tan } x \right)\left( 1 - \text{ tan } x \right)}\]
\[ = \frac{2(1 + \tan^2 x)}{\left( 1 - \tan^2 x \right)} = \frac{2\left( \sec^2 x \right)}{1 - \frac{\sin^2 x}{\cos^2 x}}\]
\[= \frac{2\left( \sec^2 x \right)\left( \cos^2 x \right)}{\cos2x} \left( \because \cos^2 x - \sin^2 x = \cos2x \right)\]
\[ = \frac{2 \times 1}{\cos2x}\]
\[ = 2\sec2x = RHS\]
\[\text{ Hence proved } .\]
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