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Prove that: (i) tan8θ−tan6θ−tan2θ=tan8θtan6θtan2θ (ii) tan15∘+tan30∘+tan15∘tan30∘=1 (iii) tan36∘+tan9∘+tan36∘tan9∘=1 (iv) tan13θ−tan9θ−tan4θ=tan13θtan9θtan4θ

Answer» Prove that:
(i) tan8θtan6θtan2θ=tan8θtan6θtan2θ
(ii) tan15+tan30+tan15tan30=1
(iii) tan36+tan9+tan36tan9=1
(iv) tan13θtan9θtan4θ=tan13θtan9θtan4θ


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