Question

Prove that
$\tan (\pi /4+\theta )\times \tan (\pi /4-\theta )=1$

Answer 1

$\tan (\frac{\pi }{4}+\theta )\times \tan (\frac{\pi }{4}-\theta )$

$\frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\tan \theta }\times \frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta }$

$\frac{1+\tan \theta }{1-\tan \theta }\times \frac{1-\tan \theta }{1+\tan \theta }=1$

$\therefore \tan (\frac{\pi }{4}+\theta )\times \tan (\frac{\pi }{4}-\theta )=1$