Question

If $\tan (\pi /4+x)+\tan (\pi /4-x)=a$, then ${\tan }^2(\pi /4+x)+{\tan }^2(\pi /4-x)=?$

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

Answer 1

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

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

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

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

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

Squaring on both sides

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

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

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