Prove that tan 8θ -tan 6θ- tan 2θ = tan 8θ tan 6θtan 2θ
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1 Answers
Manish Shankar
·2014-03-10 22:39:34
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)
- Himanshu Giria thank you sir ...
Upvote·0· Reply ·2014-03-11 02:22:33