If $(\sec A-\tan A)=x$ then prove that $\frac{1+x^{2}}{1-x^{2}}=\operatorname{cosec} A$.
Given: $\sec A-\tan A=x \quad \ldots \ldots$ (1)
We know
$\sec ^{2} A-\tan ^{2} A=1$
$\Rightarrow(\sec A+\tan A)(\sec A-\tan A)=1 \quad\left[a^{2}-b^{2}=(a-b)(a+b)\right]$
$\Rightarrow(\sec A+\tan A) x=1 \quad[$ From (1) $]$
$\Rightarrow \sec A+\tan A=\frac{1}{x} \quad \ldots(2)$
Adding (1) and (2), we get
$\sec A-\tan A+\sec A+\tan A=x+\frac{1}{x}$
$\Rightarrow 2 \sec A=\frac{x^{2}+1}{x}$
$\Rightarrow \sec A=\frac{x^{2}+1}{2 x} \quad \ldots \ldots(3)$
Subtracting (1) from (2), we get
$\sec A+\tan A-\sec A+\tan A=\frac{1}{x}-x$
$\Rightarrow 2 \tan A=\frac{1-x^{2}}{x}$
$\Rightarrow \tan A=\frac{1-x^{2}}{2 x} \quad \ldots \ldots(4)$
Dividing (3) by (4), we get
$\frac{\frac{1+x^{2}}{2 x}}{\frac{1-z^{2}}{2 x}}=\frac{\sec A}{\tan A}$
$\Rightarrow \frac{1+x^{2}}{1-x^{2}}=\frac{\frac{1}{\cos A}}{\frac{\sin A}{\cos A}}$
$\Rightarrow \frac{1+x^{2}}{1-x^{2}}=\frac{1}{\sin A}$
$\Rightarrow \frac{1+x^{2}}{1-x^{2}}=\operatorname{cosec} A$