Q(23) Prove that

 \small \tan4x = \frac{4\tan x(1-\tan^{2}x)}{1-6 \tan^{2}x+\tan^{4}x}

Answers (1)
G Gautam harsolia

We know that 
 
  tan2A=\frac{2\tan A}{1 - \tan^{2}A}

and we can write tan 4x = tan 2(2x)
So,  tan4x=\frac{2\tan 2x}{1 - \tan^{2}2x}     =   \frac{2( \frac{2\tan x}{1 - \tan^{2}x})}{1 - (\frac{2\tan x}{1 - \tan^{2}x})^{2}}      
 

                                                  =  \frac{2 (2\tan x)(1 - \tan^{2}x)}{(1-\tan x)^{2} - (4\tan^{2} x)}

                                                   =  \frac{(4\tan x)(1 - \tan^{2}x)}{(1)^{2}+(\tan^{2} x)^{2} - 2 \tan^{2} x - (4\tan^{2} x)}

                                                    =  \frac{(4\tan x)(1 - \tan^{2}x)}{1^{2}+\tan^{4} x - 6 \tan^{2} x }         = R.H.S.

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