Question
Prove that $\sec ^2 A-\operatorname{cosec}^2 A=\frac{2 \sin ^2 A-1}{\sin ^2 A \cdot \cos ^2 A}$
✓
Answer
$ \text { L.H.S }=\sec ^2 A-\operatorname{cosec}^2 A$
$ =\frac{1}{\cos ^2 A}-\frac{1}{\sin ^2 A}$
$ =\frac{\sin ^2 A-\cos ^2 A}{\cos ^2 A \cdot \sin ^2 A}$
$=\frac{\sin ^2 A-\left(1-\sin ^2 A\right)}{\sin ^2 A \cdot \cos ^2 A} \ldots\left[\begin{array}{l}\because \sin ^2 A+\cos ^2 A=1 \\ \therefore 1-\sin ^2 A=\cos ^2 A\end{array}\right]$
$=\frac{\sin ^2 A-1+\sin ^2 A}{\sin ^2 A \cdot \cos ^2 A}$
$ =\frac{2 \sin ^2 A-1}{\sin ^2 A \cdot \cos ^2 A}$
$ =\text { R.H.S }$
$\therefore \sec ^2 A-\operatorname{cosec}^2 A=\frac{2 \sin ^2 A-1}{\sin ^2 A \cdot \cos ^2 A}$
$ =\frac{1}{\cos ^2 A}-\frac{1}{\sin ^2 A}$
$ =\frac{\sin ^2 A-\cos ^2 A}{\cos ^2 A \cdot \sin ^2 A}$
$=\frac{\sin ^2 A-\left(1-\sin ^2 A\right)}{\sin ^2 A \cdot \cos ^2 A} \ldots\left[\begin{array}{l}\because \sin ^2 A+\cos ^2 A=1 \\ \therefore 1-\sin ^2 A=\cos ^2 A\end{array}\right]$
$=\frac{\sin ^2 A-1+\sin ^2 A}{\sin ^2 A \cdot \cos ^2 A}$
$ =\frac{2 \sin ^2 A-1}{\sin ^2 A \cdot \cos ^2 A}$
$ =\text { R.H.S }$
$\therefore \sec ^2 A-\operatorname{cosec}^2 A=\frac{2 \sin ^2 A-1}{\sin ^2 A \cdot \cos ^2 A}$
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