tan(A-B)=(tanA-tanB)/(1+tanAtanB)

Prove that:

tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB }

Firstly:

tan\left( \alpha +\beta \right) =\frac { tan\alpha +tan\beta }{ 1-tan\alpha tan\beta }

Also, remember that:

tan\left( -\theta \right) =-tan\theta

So:

\beta =-\phi \\ \\ tan\left( \alpha +\left( -\phi \right) \right) =\frac { tan\alpha +tan\left( -\phi \right) }{ 1-tan\alpha tan\left( -\phi \right) } \\ \\ =\frac { tan\alpha -tan\phi }{ 1+tan\alpha tan\phi } =tan\left( \alpha -\phi \right) \\ \\ \therefore \quad tan\left( A-B \right) =\frac { tanA-tanB }{ 1+tanAtanB }

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