Question
Prove the following identities:
$
\frac{\sec A-1}{\sec A+1}=\left(\frac{\sin A}{1+\cos A}\right)^2
$
$
\frac{\sec A-1}{\sec A+1}=\left(\frac{\sin A}{1+\cos A}\right)^2
$
✓
Answer
$
\begin{aligned}
& \frac{\sec \mathrm{A}-1}{\sec \mathrm{A}+1}=\left(\frac{\sin \mathrm{A}}{1+\cos \mathrm{A}}\right)^2 \\
& =\text { L.H.S. } \\
& =\frac{\sec \mathrm{A}-1}{\sec \mathrm{A}+1} \\
& =\frac{\frac{1}{\cos A}-1}{\frac{1}{\cos A}+1} \\
& =\frac{1-\cos A}{1+\cos A}
\end{aligned}
$
Multiplied by $1+\cos A$
$
\begin{aligned}
& =\frac{1-\cos ^2 A}{(1+\cos A)^2} \\
& =\frac{\sin ^2 A}{(1+\cos A)^2} \\
& =\left(\frac{\sin A}{1+\cos A}\right)^2 \\
& =\text { R.H.S }
\end{aligned}
$
Hence Proved.
\begin{aligned}
& \frac{\sec \mathrm{A}-1}{\sec \mathrm{A}+1}=\left(\frac{\sin \mathrm{A}}{1+\cos \mathrm{A}}\right)^2 \\
& =\text { L.H.S. } \\
& =\frac{\sec \mathrm{A}-1}{\sec \mathrm{A}+1} \\
& =\frac{\frac{1}{\cos A}-1}{\frac{1}{\cos A}+1} \\
& =\frac{1-\cos A}{1+\cos A}
\end{aligned}
$
Multiplied by $1+\cos A$
$
\begin{aligned}
& =\frac{1-\cos ^2 A}{(1+\cos A)^2} \\
& =\frac{\sin ^2 A}{(1+\cos A)^2} \\
& =\left(\frac{\sin A}{1+\cos A}\right)^2 \\
& =\text { R.H.S }
\end{aligned}
$
Hence Proved.
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