C-09-Q093mediumsingle_mcq1+sinθcosθ+cosθ1+sinθ=?\dfrac{1 + \sin\theta}{\cos\theta} + \dfrac{\cos\theta}{1 + \sin\theta} = ?cosθ1+sinθ+1+sinθcosθ=?a2secθ2\sec\theta2secθb2cosecθ2\operatorname{cosec}\theta2cosecθc2sinθ2\sin\theta2sinθdsecθ\sec\thetasecθব্যাখ্যাCombining over a common denominator gives (1+sinθ)2+cos2θcosθ(1+sinθ)=1+2sinθ+sin2θ+cos2θcosθ(1+sinθ)=2(1+sinθ)cosθ(1+sinθ)\dfrac{(1+\sin\theta)^2 + \cos^2\theta}{\cos\theta(1+\sin\theta)} = \dfrac{1 + 2\sin\theta + \sin^2\theta + \cos^2\theta}{\cos\theta(1+\sin\theta)} = \dfrac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)}cosθ(1+sinθ)(1+sinθ)2+cos2θ=cosθ(1+sinθ)1+2sinθ+sin2θ+cos2θ=cosθ(1+sinθ)2(1+sinθ). This simplifies to 2cosθ=2secθ\dfrac{2}{\cos\theta} = 2\sec\thetacosθ2=2secθ.