Question
1+sin(90∘−θ)−cos2(90∘−θ)cos(90∘−θ) [1+sin(90∘−θ)]=
Answer: Option A
:
A
1+sin(90∘−θ)−cos2(90∘−θ)cos(90∘−θ)[1+sin(90∘−θ)]
=1+cosθ−sin2θsinθ(1+cosθ)
=(1−sin2θ)+cosθsinθ(1+cosθ)
=cos2θ+cosθsinθ(1+cosθ)
=cosθ(cosθ+1)sinθ(1+cosθ)=cotθ
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:
A
1+sin(90∘−θ)−cos2(90∘−θ)cos(90∘−θ)[1+sin(90∘−θ)]
=1+cosθ−sin2θsinθ(1+cosθ)
=(1−sin2θ)+cosθsinθ(1+cosθ)
=cos2θ+cosθsinθ(1+cosθ)
=cosθ(cosθ+1)sinθ(1+cosθ)=cotθ
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