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Question

# If tan θ2=√1−e1+e. tan α2, then cos α =

A

1e cos (cos θ+e)

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B

1+e cos θcos θe

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C

1e cos θcos θe

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D

cos θe1e cos θ

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Solution

## The correct option is D cos θ−e1−e cos θ Given: tan θ2=√1−e1+e. tan α2⇒ tan θ2tan α2=√1−e1+e Squaring both sides, we get, tan2θ2tan2α2=1−e1+e⇒ tan2 α2 (1−e)=tan2 θ2(1+e)⇒ sin2α2cos2α2(1−e)=sin2θ2cos2θ2(1+e)⇒ 12(1−cos α)12(1+cos α)(1−e) =12(1−cos θ)12(1+cosθ)(1+e)⇒ (1−cos α)(1+cosθ)(1−e)=(1+cos α)(1−cos α)(1+e)⇒ (1+cos θ)(1−e)−cos α(1+cos θ)(1−e)=(1−cos θ)(1+e)+cos α(1−cos θ)(1+e)⇒ cos α (1+cos θ)(1−e)−(1−cos θ)(1+e)=(1+cos θ)(1−e)−(1−cos θ)(1+e) ⇒ cos α=2 cos θ−2e2−2 cos θ=cos θ−e1−e cos θ

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