Question
Prove. $\tan^2A - \sin^2A = \tan^2A \sin^2A$
✓
Answer
$\text { LHS }=\tan ^2 A-\sin ^2 A$
$=\frac{\sin ^2 A}{\cos ^2 A}-\sin ^2 A=\frac{\sin ^2 A\left(1-\cos ^2 A\right)}{\cos ^2 A}$
$=\frac{\sin ^2 A}{\cos ^2 A} \cdot \sin ^2 A=\tan ^2 A \cdot \sin ^2 A=\text { RHS }$
$=\frac{\sin ^2 A}{\cos ^2 A}-\sin ^2 A=\frac{\sin ^2 A\left(1-\cos ^2 A\right)}{\cos ^2 A}$
$=\frac{\sin ^2 A}{\cos ^2 A} \cdot \sin ^2 A=\tan ^2 A \cdot \sin ^2 A=\text { RHS }$
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