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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove : $\tan 4x=\large\frac{4\tan x(1-\tan^2x)}{1-6\tan^2x+\tan^4x}$

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  • $\tan 2x=\large\frac{2\tan x}{1-\tan^2x}$
L.H.S
$\tan 2(2x)=\large\frac{2\tan 2x}{1-\tan^22x}$
$\Rightarrow \large\frac{2(\Large\frac{2\tan x}{1-\tan^2x})}{1-(\Large\frac{2\tan x}{1-\tan^2x})^2}$
$\Rightarrow \large\frac{\Large\frac{4\tan x}{1-\tan^2x}}{\Large\frac{(1-\tan^2x)^2-4\tan^2x}{(1-\tan^2x)^2}}$
$\Rightarrow \large\frac{4\tan x}{1-\tan^2x}\times \large\frac{(1-\tan^2x)^2}{(1-\tan^2x)^2-4\tan^2x}$
$\Rightarrow \large\frac{4\tan x(1-\tan^2x)}{1+\tan^4x-2\tan^2x-4\tan^2x}$
$\Rightarrow \large\frac{4\tan x(1-\tan^2x)}{1+\tan^4x-6\tan^2x}$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 
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