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Question

# If (sec A – tan A) = x then prove that $\frac{1+{x}^{2}}{1-{x}^{2}}$= cosec A.

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Solution

## Given: $\mathrm{sec}A-\mathrm{tan}A=x.....\left(1\right)$ We know ${\mathrm{sec}}^{2}A-{\mathrm{tan}}^{2}A=1\phantom{\rule{0ex}{0ex}}⇒\left(\mathrm{sec}A+\mathrm{tan}A\right)\left(\mathrm{sec}A-\mathrm{tan}A\right)=1\left[{a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)\right]\phantom{\rule{0ex}{0ex}}⇒\left(\mathrm{sec}A+\mathrm{tan}A\right)x=1\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}A+\mathrm{tan}A=\frac{1}{x}.....\left(2\right)$ Adding (1) and (2), we get $\mathrm{sec}A-\mathrm{tan}A+\mathrm{sec}A+\mathrm{tan}A=x+\frac{1}{x}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{sec}A=\frac{{x}^{2}+1}{x}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sec}A=\frac{{x}^{2}+1}{2x}.....\left(3\right)$ Subtracting (1) from (2), we get $\mathrm{sec}A+\mathrm{tan}A-\mathrm{sec}A+\mathrm{tan}A=\frac{1}{x}-x\phantom{\rule{0ex}{0ex}}⇒2\mathrm{tan}A=\frac{1-{x}^{2}}{x}\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}A=\frac{1-{x}^{2}}{2x}.....\left(4\right)$ Dividing (3) by (4), we get $\frac{\frac{1+{x}^{2}}{2x}}{\frac{1-{x}^{2}}{2x}}=\frac{\mathrm{sec}A}{\mathrm{tan}A}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\frac{\frac{1}{\mathrm{cos}A}}{\frac{\mathrm{sin}A}{\mathrm{cos}A}}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\frac{1}{\mathrm{sin}A}\phantom{\rule{0ex}{0ex}}⇒\frac{1+{x}^{2}}{1-{x}^{2}}=\mathrm{cosec}A$

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