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

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प्रश्न

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

संख्यात्मक

उत्तर

\[LHS = \frac{\cos x}{1 - \sin x}\]

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

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

\[ = \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\]

On dividing the numerator and denominator by

\[\cos\frac{x}{2}\]

, we get

\[= \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}\]

\[ = \tan\left( \frac{\pi}{4} + \frac{x}{2} \right) = RHS\]

\[\text{ Hence proved } .\]

 
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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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अध्याय 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.1 [पृष्ठ २८]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 8 | पृष्ठ २८

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