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

Answer

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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.

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.

Last updated date: 26th Sep 2023

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