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Prove :$\large\frac{\tan(\Large\frac{\pi}{4}\normalsize +x)}{\tan(\Large\frac{\pi}{4}\normalsize -x)}=\big(\large\frac{1+\tan x}{1-\tan x}\big)^2$

1 Answer

Toolbox:
  • $\tan (A+B)=\large\frac{\tan A+\tan B}{1-\tan A.\tan B}$
  • $\tan (A-B)=\large\frac{\tan A-\tan B}{1+\tan A.\tan B}$
  • $\tan \large\frac{\pi}{4}$$=1$
L.H.S
$\large\frac{\tan(\Large\frac{\pi}{4}\normalsize +x)}{\tan(\Large\frac{\pi}{4}\normalsize -x)}=\large\frac{\Large\frac{\tan (\pi/4)+\tan x}{1-\tan(\pi/4).\tan x}}{\Large\frac{\tan (\pi/4)-\tan x}{1+\tan(\pi/4).\tan x}}$
$\Rightarrow \large\frac{\Large\frac{1+\tan x}{1-\tan x}}{\Large\frac{1-\tan x}{1+\tan x}}$
$\Rightarrow \large\frac{1+\tan x}{1-\tan x}$$\times \large\frac{1+\tan x}{1-\tan x}$
$\Rightarrow \big(\large\frac{1+\tan x}{1-\tan x}\big)^2$=R.H.S
Hence proved
answered Apr 17, 2014 by sreemathi.v
 

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