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Prove That: Cos 2 X 1 + Sin 2 X = Tan ( π 4 − X ) - Mathematics

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Question

Prove that:  \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]

 
Numerical
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Solution

\[LHS = \frac{\cos2x}{1 + \sin2x}\]

\[= \frac{\cos^2 x - \sin^2 x}{\sin^2 x + \cos^2 x + 2\text{ sin } x \times \text{ cos } x} \left[ \because \cos2x = \cos^2 x - \sin^2 x \text{ and }  \sin^2 x + \cos^2 x = 1 \right]\]

\[ = \frac{\left( \text{ cos } x - \text{ sin } x \right)\left( \text{ cos } x + \text{ sin } x \right)}{\left( \text{ cos } x + \text{ sin } x \right)^2} \left[ \because a^2 - b^2 = \left( a + b \right)\left( a - b \right) \right]\]

\[ = \frac{\text{ cos } x - \text{ sin } x}{\text{ cos } x + \text{ sin } x}\]

On dividing the numerator and denominator by cos x, we get

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

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

\[\text{ Hence proved } .\]

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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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RD Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 7 | Page 28

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