If (sec A – tan A) = x then prove that


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

$\frac{\frac{1+x^{2}}{2 x}}{\frac{1-z^{2}}{2 x}}=\frac{\sec A}{\tan A}$

$\Rightarrow \frac{1+x^{2}}{1-x^{2}}=\frac{\frac{1}{\cos A}}{\frac{\sin A}{\cos A}}$

$\Rightarrow \frac{1+x^{2}}{1-x^{2}}=\frac{1}{\sin A}$

$\Rightarrow \frac{1+x^{2}}{1-x^{2}}=\operatorname{cosec} A$


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