Question
Prove.$\frac{1}{\sec A+\tan A}=\sec A-\tan A$
✓
Answer
$
\begin{aligned}
& \text { LHS }=\frac{1}{\sec A +\tan A } \\
& =\frac{1}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}} \\
& =\frac{1}{\frac{1+\sin A }{\cos A }} \\
& =\frac{\cos A }{1+\sin A } \times \frac{1-\sin A }{1+\sin A } \\
& =\frac{\cos A(1-\sin A)}{(1)^2-\sin ^2 A} \\
& =\frac{\cos A(1-\sin A)}{\cos ^2 A} \\
& =\frac{1}{\cos A}-\frac{\sin A}{\cos A} \\
& =\sec A -\tan A \\
& \text { LHS }=\text { RHS }
\end{aligned}
$Hence proved.q
\begin{aligned}
& \text { LHS }=\frac{1}{\sec A +\tan A } \\
& =\frac{1}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}} \\
& =\frac{1}{\frac{1+\sin A }{\cos A }} \\
& =\frac{\cos A }{1+\sin A } \times \frac{1-\sin A }{1+\sin A } \\
& =\frac{\cos A(1-\sin A)}{(1)^2-\sin ^2 A} \\
& =\frac{\cos A(1-\sin A)}{\cos ^2 A} \\
& =\frac{1}{\cos A}-\frac{\sin A}{\cos A} \\
& =\sec A -\tan A \\
& \text { LHS }=\text { RHS }
\end{aligned}
$Hence proved.q
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