KeerthanaPosted on sinθ1+cosθ+1+cosθsinθ=?\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=?1+cosθsinθ+sinθ1+cosθ=? a2 sin θ b2 cos θ c2 sec θ d2 cosec θ Answer : Option DExplanation : sinθ1+cosθ+1+cosθsinθ=sin2θ+(1+cosθ)2(1+cosθ)sinθ\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=\frac{\sin ^{2} \theta+(1+\cos \theta)^{2}}{(1+\cos \theta) \sin \theta}1+cosθsinθ+sinθ1+cosθ=(1+cosθ)sinθsin2θ+(1+cosθ)2 =sin2θ+1+2cosθ+cos2θ(1+cosθ)sinθ=\frac{\sin ^{2} \theta+1+2 \cos \theta+\cos ^{2} \theta}{(1+\cos \theta) \sin \theta}=(1+cosθ)sinθsin2θ+1+2cosθ+cos2θ =sin2θ+cos2θ+1+2cosθ(1+cosθ)sinθ=2+2cosθ(1+cosθ)sinθ=\frac{\sin ^{2} \theta+\cos ^{2} \theta+1+2 \cos \theta}{(1+\cos \theta) \sin \theta}=\frac{2+2 \cos \theta}{(1+\cos \theta) \sin \theta}=(1+cosθ)sinθsin2θ+cos2θ+1+2cosθ=(1+cosθ)sinθ2+2cosθ =2(1+cosθ)(1+cosθ)sinθ=2sinθ=\frac{2(1+\cos \theta)}{(1+\cos \theta) \sin \theta}=\frac{2}{\sin \theta}=(1+cosθ)sinθ2(1+cosθ)=sinθ2 = 2 cosec θ Rate This:NaN / 5 - 1 votesAdd comment
