Q (7)  Prove the following

\small \frac{\tan \left ( \frac{\pi }{4}+x \right )}{\tan \left ( \frac{\pi }{4} -x\right )} = \left ( \frac{1+\tan x}{1-\tan x} \right )^{2}

Answers (1)
S safeer

As we know that

(\tan (A +B ) = \frac {\tan A + \tan B}{1- \tan A\tan B})     and   \tan (A-B) = \frac {\tan A - \tan B }{1+ \tan A \tan B}

So, by using these identities 

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

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