Prove that: (i) tan8θ−tan6θ−tan2θ=tan8θtan6θtan2θ (ii) tan15 – InterviewSolution

1.	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) $t a n 8 θ - t a n 6 θ - t a n 2 θ = t a n 8 θ t a n 6 θ t a n 2 θ$ (ii) $t a n 15^{\circ} + t a n 30^{\circ} + t a n 15^{\circ} t a n 30^{\circ} = 1$ (iii) $t a n 36^{\circ} + t a n 9^{\circ} + t a n 36^{\circ} t a n 9^{\circ} = 1$ (iv) $t a n 13 θ - t a n 9 θ - t a n 4 θ = t a n 13 θ t a n 9 θ t a n 4 θ$

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