Question

If $\tan \left(\frac{\pi }{4}+x\right)+\tan \left(\frac{\pi }{4}-x\right)=a$, then ${\tan }^2\left(\frac{\pi }{4}+x\right)+{\tan }^2\left(\frac{\pi }{4}-x\right)=$

• A. $a^2+1$
• B. $a^2+2$
• C. $a^2-2$
• D. none of these

Answer 1

The correct option is C $a^2-2$

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

Square both sides,

$\Rightarrow {[\tan (\frac{\pi }{4}+x)+\tan (\frac{\pi }{4}-x)]}^2=a^2$

$\Rightarrow {\tan }^2(\frac{\pi }{4}+x)+{\tan }^2(\frac{\pi }{4}-x)+2\tan (\frac{\pi }{4}+x)\tan (\frac{\pi }{4}-x)=a^2$

$\Rightarrow {\tan }^2(\frac{\pi }{4}+x)+{\tan }^2(\frac{\pi }{4}-x)=a^2-2\tan (\frac{\pi }{4}+x)\tan (\frac{\pi }{4}-x)⟶\left(1\right)$

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

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

So, $\tan \pi /4=1$

So, $\tan (\frac{\pi }{4}+x)=\frac{1+\tan x}{1-\tan x}$

$\tan (\frac{\pi }{4}-x)=\frac{1-\tan x}{1+\tan x}$

$\Rightarrow \tan (\frac{\pi }{4}+x)\tan (\frac{\pi }{4}-x)=1$

So, ${\tan }^2(\frac{\pi }{4}+x)+{\tan }^2(\frac{\pi }{4}-x)=a^2-2\left(1\right)=a^2-2$.