What is the value of tanθ2?\tan \frac{\theta}{2}?tan2θ? Acosθ1−sinθ\frac{\cos \theta}{1 - \sin \theta}1−sinθcosθ Bsinθ1−cosθ\frac{\sin \theta}{1 - \cos \theta}1−cosθsinθ Ccosθ1−cosθ\frac{\cos \theta}{1 - \cos \theta}1−cosθcosθ Dsinθ1+cosθ\frac{\sin \theta}{1 + \cos \theta}1+cosθsinθ Show answer & explanationCorrect answer: DExplanationThe correct answer is Option 4 i.e. sinθ1+cosθ\frac{\sin \theta}{1 + \cos \theta}1+cosθsinθ. sin2A = 2sinA.cosA and 2cos2A - 1 = cos2A tanθ2\tan \frac{\theta}{2}tan2θ = sinθ2cosθ2{sin {\theta \over 2}} \over {cos {\theta \over 2}}cos2θsin2θ ⇒ tanθ2\tan \frac{\theta}{2}tan2θ = 2sinθ22cosθ2{2sin {\theta \over 2}} \over {2cos {\theta \over 2}}2cos2θ2sin2θ ×cosθ2cosθ2{cos{\theta \over 2}} \over {cos {\theta \over 2}}cos2θcos2θ ⇒ tanθ2\tan \frac{\theta}{2}tan2θ = 2sinθ2.cosθ22cos2θ2{2sin {\theta \over 2}}.cos {\theta \over 2} \over {2cos^2 {\theta \over 2}}2cos22θ2sin2θ.cos2θ ⇒ tanθ2\tan \frac{\theta}{2}tan2θ = sinθ2cos2θ2−1+1{sin\theta} \over {2cos^2 {\theta \over 2}-1+1}2cos22θ−1+1sinθ ⇒ tanθ2\tan \frac{\theta}{2}tan2θ = sinθ(1+cosθ){sin \theta} \over{(1 + cos \theta)}(1+cosθ)sinθ