Question

# Prove that $\tan \left( {\dfrac{\pi }{4} + \theta } \right) \times \tan \left( {\dfrac{\pi }{4} - \theta } \right) = 1$

Hint: Use the formula of $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ and $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$

Therefore, if we put the above formulas to the given equation, it becomes,
$\tan \left( {\dfrac{\pi }{4} + \theta } \right) \times \tan \left( {\dfrac{\pi }{4} - \theta } \right)$
$\dfrac{{\tan \dfrac{\pi }{4} + \tan \theta }}{{1 - \tan \dfrac{\pi }{4}\tan \theta }} \times \dfrac{{\tan \dfrac{\pi }{4} - \tan \theta }}{{1 + \tan \dfrac{\pi }{4}\tan \theta }}$
The next step is to equate$\tan \left( {\dfrac{\pi }{4}} \right) = 1$,
$\dfrac{{1 + \tan \theta }}{{1 - \tan \theta }} \times \dfrac{{1 - \tan \theta }}{{1 + \tan \theta }}$
Cancel out the common terms in denominator with numerators, we are left with,
$1 \times 1 = 1$
Hence Proved!

Note: Use the necessary formula to solve the expression. Instead of trying to solve it further, it is simpler to cancel out the terms.