Question

$\frac{tanA}{secA-1}=\frac{tanA+secA+1}{secA-1+tanA}$

Answer 1

$\Rightarrow$ Prove

$\frac{\tan A}{\sec A-1}=\frac{\tan A+\sec A-1}{\sec A-1+\tan A}$

We Know $1+{\tan }^{2}A={\sec }^{2}A$

${\tan }^{2}A=\sec A-1$

${\tan }^{2}A=(\sec A+1)(\sec A-1)$

$\frac{\tan A}{\sec A-1}=\frac{\sec A+1}{\tan A}$

$\frac{\tan A+\sec A+1}{\sec A-1+\tan A}=\frac{\tan A}{\sec A-1}=\frac{\sec A+1}{\tan A}$

[Theorem on equal Ratio]

Hence prove.