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If $\cos θ_{1} = 2 \cos θ_{2},$ then $\tan \frac{θ_{1} - θ_{2}}{2} \tan \frac{θ_{1} + θ_{2}}{2}$

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detailed solution

Correct option is B

tan⁡θ1−θ22tan⁡θ1+θ22=2sin⁡θ1+θ22sin⁡θ1−θ222cos⁡θ1+θ22cos⁡θ1−θ22=cos⁡θ2−cos⁡θ1cos⁡θ1+cos⁡θ2=−13

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If cos⁡θ1=2cos⁡θ2,then tan⁡θ1−θ22tan⁡θ1+θ22

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If $\cos θ_{1} = 2 \cos θ_{2},$ then $\tan \frac{θ_{1} - θ_{2}}{2} \tan \frac{θ_{1} + θ_{2}}{2}$