Question

prove tan(alpha-pi) = tan(alpha)

Answer 1

Using
\[\tan(A \pm B) = \frac{\tan(A)\pm \tan(B)}{1-\tan(A)\tan(B)}\]
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\[\frac{\tan(\alpha)- \tan(\pi)}{1+\tan(\alpha)\tan(\pi)}\]
Note that \(tan(\pi)=0\)

Answer 2

That was a lot easier than I thought it would be! I ahd forgotten that tan(pi)= 0! thanks a lot!

Answer 3

yw