Question

tan4Θ+tan2Θ/1-tan4Θtan2Θ
a. tan6Θ b. cot6Θ c. tan4Θ d.cot4Θ

Answer 1

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Answer 2

\[\tan(\alpha + \beta)=\frac{ \tan(\alpha)+\tan(\beta) }{ 1-\tan(\alpha)\tan(\beta) }\]How about\[\alpha=4\theta\]and \[\beta=2\theta\]Then the solution is\[\tan(6\theta)\]