मराठी

Prove That: 1 − Cos 2 X + Sin 2 X 1 + Cos 2 X + Sin 2 X = Tan X

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प्रश्न

Prove that:  \[\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x\]

 
संख्यात्मक
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उत्तर

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

\[= \frac{2 \sin^2 x + \sin2x}{2 \cos^2 x + \sin2x} \left[ \because 2 \sin^2 x = 1 - \cos2x and 2 \cos^2 x = 1 + \cos2x \right]\]

\[ = \frac{2 \sin^2 x + 2\text{ sin } x \text{ cos } x}{2 \cos^2 x + 2\text{ sin } x \text{ cos } x} \left( \because \sin2x = 2\text{ sin } x \text{ cos } x \right)\]

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

\[ = tan\theta = RHS\]

\[\text{ Hence proved }.\]

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

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 5 | पृष्ठ २८

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