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Prove That: Tan 2 2 X − Tan 2 X 1 − Tan 2 2 X Tan 2 X = Tan 3 X Tan X

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प्रश्न

Prove that:
\[\frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} = \tan 3x \tan x\]

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उत्तर

\[\text{ LHS }= \frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x}\]
\[ = \frac{(\tan2x + \tan x)(\tan2x - \tan x)}{1 - \tan^2 2x \tan^2 x} \left\{\text{ Using }A^2 - B^2 = \left( A + B \right)\left( A - B \right) \right\}\]
\[ \tan3x=\tan(2x+x)\text{ and }\tan x=\tan(2x-x) .\]
\[ = \frac{\tan3 x\left( 1 - \tan2x \tan x \right) \times \tan x\left( 1 + \tan2x \tan x \right)}{1 - \tan^2 2x \tan^2 x} \left[ \because \tan 2x + \tan x = \tan3x\left( 1 - \tan2x \tanx \right) \tan 2x - \tanx = \tanx\left( 1 + \tan2x \tanx \right) \right]\]
\[ = \frac{\tan3x \tan x (1 - \tan^2 2x \tan^2 x)}{1 - \tan^2 2x \tan^2 x}\]
\[ = \tan3 x\tan x\]
= RHS 
Hence proved . 

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अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [पृष्ठ २०]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 18 | पृष्ठ २०

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