KeerthanaPosted on The value of tan2θ1+tan2θ+cot2θ1+cot2θ\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}+\frac{\cot ^{2} \theta}{1+\cot ^{2} \theta}1+tan2θtan2θ+1+cot2θcot2θ is equal to a0 b1 c2 d3 Answer : Option BExplanation : tan2θ1+tan2θ+cot2θ1+cot2θ\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}+\frac{\cot ^{2} \theta}{1+\cot ^{2} \theta}1+tan2θtan2θ+1+cot2θcot2θ =tan2θsec2θ+cot2θcosec2θ=\frac{\tan ^{2} \theta}{\sec ^{2} \theta}+\frac{\cot ^{2} \theta}{\operatorname{cosec}^{2} \theta}=sec2θtan2θ+cosec2θcot2θ =sin2θcos2θ×cos2θ+cos2θsin2θ×sin2θ=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \cos ^{2} \theta+\frac{\cos ^{2} \theta}{\sin ^{2} \theta} \times \sin ^{2} \theta=cos2θsin2θ×cos2θ+sin2θcos2θ×sin2θ =sin2θ+cos2θ=1=\sin ^{2} \theta+\cos ^{2} \theta=1=sin2θ+cos2θ=1 Rate This:NaN / 5 - 1 votesAdd comment