Question
Prove the following identity:
$
\frac{\sec A-1}{\sec A+1}=\frac{\sin ^2 A}{(1+\cos A)^2}
$
$
\frac{\sec A-1}{\sec A+1}=\frac{\sin ^2 A}{(1+\cos A)^2}
$
✓
Answer
$
\begin{aligned}
& \text { LHS }=\frac{\sec A-1}{\sec A+1} \\
& =\frac{\frac{1}{\cos A}-1}{\frac{1}{\cos A}+1}=\frac{1-\cos A}{1+\cos A} \\
& =\frac{1-\cos A}{1+\cos A} \times \frac{1+\cos A}{1+\cos A} \\
& =\frac{1-\cos ^2 A}{(1+\cos A)^2} \\
& =\frac{\sin ^2 A}{(1+\cos A)^2} \quad\left(\because 1-\cos ^2 A=\sin ^2 A\right)
\end{aligned}
$
\begin{aligned}
& \text { LHS }=\frac{\sec A-1}{\sec A+1} \\
& =\frac{\frac{1}{\cos A}-1}{\frac{1}{\cos A}+1}=\frac{1-\cos A}{1+\cos A} \\
& =\frac{1-\cos A}{1+\cos A} \times \frac{1+\cos A}{1+\cos A} \\
& =\frac{1-\cos ^2 A}{(1+\cos A)^2} \\
& =\frac{\sin ^2 A}{(1+\cos A)^2} \quad\left(\because 1-\cos ^2 A=\sin ^2 A\right)
\end{aligned}
$
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$=a\left(\frac{1}{t^2}+1\right)=\frac{a\left(t^2+1\right)}{t^2}$
$\text { Now } \frac{1}{ SP }+\frac{1}{ SQ }=\frac{1}{a\left(t^2+1\right)}+\frac{1 \times t^2}{a\left(t^2+1\right)}$
$=\frac{\left(1+t^2\right)}{a\left(t^2+1\right)}$
$\frac{1}{ SP }+\frac{1}{ SQ }=\frac{1}{a} .$
→$=a\left(\frac{1}{t^2}+1\right)=\frac{a\left(t^2+1\right)}{t^2}$
$\text { Now } \frac{1}{ SP }+\frac{1}{ SQ }=\frac{1}{a\left(t^2+1\right)}+\frac{1 \times t^2}{a\left(t^2+1\right)}$
$=\frac{\left(1+t^2\right)}{a\left(t^2+1\right)}$
$\frac{1}{ SP }+\frac{1}{ SQ }=\frac{1}{a} .$
Find the volume of a cone whose slant height is $17\ cm$ and radius of base is $8\ cm.$
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