Trigonometric Function (1 Viewer)

tashharvey · Aug 21, 2014

I need help with the question:

tan ( a + b + c ) =

tan a + tan b + tan c - tan a tan b tan c
1 - tan a tan b - tan b tan c - tan c tan a

It asks to prove it?

Drongoski · Aug 21, 2014

Apply the identity: tan(A + B) = (tan A + tan B)/(1 - tan A x tan B) twice.

$tan(A+B+C) = tan(A + [B+C]) = \frac{tan A + tan[B+C]}{1 - tan A \times tan[B+C]} = \frac {tanA + \frac {tan B + tan C}{1 - tan B \dot tan C}}{1 - tan A \cdot \frac {tan B + tan C}{1 - tan B \cdot tan C}}$

The formula follows from this.

Trigonometric Function (1 Viewer)

tashharvey

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Drongoski

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