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If $\cos (θ + ϕ) = m \cos (θ - ϕ),$ then $\tan θ$ is equal to

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\frac{1 + m}{1 - m} \tan ϕ

\frac{1 - m}{1 + m} \tan ϕ

\frac{1 - m}{1 + m} c o t ϕ

\frac{1 + m}{1 - m} s e c ϕ

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detailed solution

Correct option is C

We have,

$\begin{array}{l} \cos (θ + ϕ) = m \cos (θ - ϕ) \\ \Rightarrow \frac{1}{m} = \frac{\cos (θ - ϕ)}{\cos (θ + ϕ)} \end{array}$

$\begin{array}{l} \Rightarrow \frac{1 + m}{1 - m} = \frac{2 \cos θ \cos ϕ}{2 \sin θ \sin ϕ} \\ \Rightarrow \tan θ \tan ϕ = \frac{1 - m}{1 + m} \Rightarrow \tan θ = \frac{1 - m}{1 + m} \cot ϕ \end{array}$

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If cos⁡(θ+ϕ)=mcos⁡(θ−ϕ),then tan⁡θ is equal to