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If $\cos θ = \cos α \cos β,$ then $\tan (\frac{θ + α}{2}) \tan (\frac{θ - α}{2})$ is equal to

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a

tan2⁡α2

b

tan2⁡β2

c

tan2⁡θ2

d

cot2⁡β2

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detailed solution

Correct option is B

We have,tan⁡θ+α2tan⁡θ−α2=2sin⁡θ+α2sin⁡θ−α22cos⁡θ+α2cos⁡θ−α2=−(cos⁡θ−cos⁡α)cos⁡θ+cos⁡α=−(cos⁡αcos⁡β−cos⁡α)cos⁡αcos⁡β+cos⁡α=−cos⁡α(cos⁡β−1)cos⁡α(cos⁡β+1)=1−cos⁡β1+cos⁡β=tan2⁡β2

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