Question

$\tan (\frac{\pi }{4}+x)+\tan (\frac{\pi }{4}-x)=2$, then $x=$

• A. $n\pi \pm \frac{\pi }{4}$
• B. $2n\pi \pm \frac{\pi }{4}$
• C. $n\pi$
• D. $(2n+1)\frac{\pi }{4}$

Answer 1

The correct option is C $n\pi$

Given $\tan (\frac{\pi }{4}+x)+\tan (\frac{\pi }{4}-x)=2$

$⟹\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}+\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}=2$

$⟹\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}=2$

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

$⟹\frac{2(1+{\tan }^{2}x)}{1-{\tan }^{2}x}=2$ $(\because (a+b{)}^2+(a-b{)}^2=2(a^2+b^2)$ and $(a+b)(a-b)=a^2-b^2)$

$⟹\frac{1+{\tan }^{2}x}{1-{\tan }^{2}x}=1$

$⟹1-{\tan }^{2}x=1+{\tan }^{2}x$

$⟹{\tan }^{2}x=0={\tan }^{2}0$

$⟹x=n\pi$