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

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−13

-1

answer is B.

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Detailed Solution

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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