trigonometry

Prove that tan 8θ -tan 6θ- tan 2θ = tan 8θ tan 6θtan 2θ

1 Answers

1357
Manish Shankar ·

tan(8θ+(-6θ)+(-2θ))=0

tan(A+B+C)=0

=> tanA+tanB+tanC=tanAtanBtanC

So we have

tan(8θ)+tan(-6θ)+tan(-2θ)=tan(8θ).tan(-6θ).tan(-2θ)

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