Question
Prove that $\frac{1+\sec A}{\sec A}=\frac{\sin ^2 A}{1-\cos A}$.
✓
Answer
$
\begin{aligned}
LHS\quad\frac{1+\sec A}{\sec A} & =\frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}} \\
& =\frac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}} \\
& =\frac{\cos A+1}{\cos A} \times \frac{\cos A}{1} \\
& =1+\cos A \\
& =1+\cos A \times\left(\frac{1-\cos A}{1-\cos A}\right) \\
& =\frac{(1)^2-(\cos A)^2}{1-\cos A} \\
& =\frac{1-\cos { }^2 A}{1-\cos A} \\
& =\frac{\sin ^2 A}{1-\cos A}=R H S
\end{aligned}
$
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