Question
Prove the following identity :
$
\sec ^4 A-\sec ^2 A=\frac{\sin ^2 A}{\cos ^4 A}
$
$
\sec ^4 A-\sec ^2 A=\frac{\sin ^2 A}{\cos ^4 A}
$
✓
Answer
$
\begin{aligned}
& \sec ^4 A-\sec ^2 A=\frac{1}{\cos ^4 A}-\frac{1}{\cos ^2 A} \\
& =\frac{1-\cos ^2 A}{\cos ^4 A} \\
& =\frac{\sin ^2 A}{\cos ^4 A} \quad\left[\because \sin ^2 A=1-\cos ^2 A\right]
\end{aligned}
$
\begin{aligned}
& \sec ^4 A-\sec ^2 A=\frac{1}{\cos ^4 A}-\frac{1}{\cos ^2 A} \\
& =\frac{1-\cos ^2 A}{\cos ^4 A} \\
& =\frac{\sin ^2 A}{\cos ^4 A} \quad\left[\because \sin ^2 A=1-\cos ^2 A\right]
\end{aligned}
$
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