C-09-Q103mediumsingle_mcq(tanθ+secθ)2=?(\tan\theta + \sec\theta)^2 = ?(tanθ+secθ)2=?a1+sinθ1−sinθ\dfrac{1 + \sin\theta}{1 - \sin\theta}1−sinθ1+sinθb1−sinθ1+sinθ\dfrac{1 - \sin\theta}{1 + \sin\theta}1+sinθ1−sinθc111d000ব্যাখ্যাExpanding, (tanθ+secθ)2=(sinθ+1cosθ)2=(1+sinθ)2cos2θ=(1+sinθ)21−sin2θ=(1+sinθ)2(1+sinθ)(1−sinθ)=1+sinθ1−sinθ(\tan\theta+\sec\theta)^2 = \left(\dfrac{\sin\theta+1}{\cos\theta}\right)^2 = \dfrac{(1+\sin\theta)^2}{\cos^2\theta} = \dfrac{(1+\sin\theta)^2}{1-\sin^2\theta} = \dfrac{(1+\sin\theta)^2}{(1+\sin\theta)(1-\sin\theta)} = \dfrac{1+\sin\theta}{1-\sin\theta}(tanθ+secθ)2=(cosθsinθ+1)2=cos2θ(1+sinθ)2=1−sin2θ(1+sinθ)2=(1+sinθ)(1−sinθ)(1+sinθ)2=1−sinθ1+sinθ.