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