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Prove That: Tan ( π 4 + X ) + Tan ( π 4 − X ) = 2 Sec 2 X

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Question

Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]

 
Numerical
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Solution

\[LHS = \tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right)\]

\[ = \frac{\tan\frac{\pi}{4} + \text{ tan } x}{1 - \tan\frac{\pi}{4}\text{ tan } x} + \frac{\tan\frac{\pi}{4} - \text{ tan } x}{1 + \tan\frac{\pi}{4}\text{ tan } x} \left[ \because \tan\left( A + B \right) = \frac{\text{ tan } A + \text{ tan } B}{1 - \text{ tan } A\text{ tan } B} \text{ and }  \tan\left( A - B \right) = \frac{\text{ tan }A - \text{ tan } B}{1 + \text{ tan } A\text{ tan } B} \right]\]

\[= \frac{1 + \text{ tan } x}{1 - \text{ tan } x} + \frac{1 - \text{ tan } x}{1 + \text{ tan } x}\]

\[ = \frac{\left( 1 + \text{ tan } x \right)^2 + \left( 1 - \text{ tan } x \right)^2}{\left( 1 + \text{ tan } x \right)\left( 1 - \text{ tan } x \right)}\]

\[ = \frac{2(1 + \tan^2 x)}{\left( 1 - \tan^2 x \right)} = \frac{2\left( \sec^2 x \right)}{1 - \frac{\sin^2 x}{\cos^2 x}}\]

\[= \frac{2\left( \sec^2 x \right)\left( \cos^2 x \right)}{\cos2x} \left( \because \cos^2 x - \sin^2 x = \cos2x \right)\]

\[ = \frac{2 \times 1}{\cos2x}\]

\[ = 2\sec2x = RHS\]

\[\text{ Hence proved }  .\]

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [Page 28]

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R.D. Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 22 | Page 28

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