tan2 θ – sin2 θ = tan2 θ sin2 θ
In this question, we can start from LHS and convert tan θ in terms of sinθ and cosθ
\begin{aligned} &=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}-\sin ^{2} \theta \\ &=\frac{\sin ^{2} \theta-\sin ^{2} \theta \cos ^{2} \theta}{\cos ^{2} \theta} \\ &=\frac{\sin ^{2} \theta\left(1-\cos ^{2} \theta\right)}{\cos ^{2} \theta}=\sin ^{2} \theta \times \frac{\sin ^{2} \theta}{\cos ^{2} \theta} \\ &\sin ^{2} \theta \times \tan ^{2} \theta=\tan ^{2} \theta \sin ^{2} \theta=\text { R.H.S. } \end{aligned}
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