plz show that [tex] \tan( \frac{\pi}{4} - \alpha ) \: \tan( \frac{\pi}{4} + \alpha ) = 1[/tex]​

Solution:The formula for tan(A+B) and tan(A-B) are:[tex]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)}[/tex]The left hand side of the given expression is:[tex]tan(\frac{\pi}{4}-\alpha ) tan(\frac{\pi}{4}+\alpha )[/tex]Using the formula above and value of tan(π/4) = 1, we can expand this expression as:[tex]tan(\frac{\pi}{4}-\alpha ) tan(\frac{\pi}{4}+\alpha )\\\\ = \frac{tan(\frac{\pi}{4} )-tan(\alpha)}{1+tan(\frac{\pi}{4} )tan(\alpha)} \times \frac{tan(\frac{\pi}{4} )+tan(\alpha)}{1-tan(\frac{\pi}{4} )tan(\alpha)}\\\\ = \frac{1-tan(\alpha)}{1+tan(\alpha)} \times \frac{1+tan(\alpha)}{1-tan(\alpha)}\\\\ = 1 \\\\ = R.H.S[/tex]Thus, the left hand side is proved to be equal to right hand side.