Question

Show that: $\mathrm{secA}(1-\mathrm{sinA})(\mathrm{secA}+\mathrm{tanA})=1$

Answer 1

Prove that $\mathrm{secA}(1-\mathrm{sinA})(\mathrm{secA}+\mathrm{tanA})=1$

Let us consider LHS, first;

$\mathrm{L}.\mathrm{H}.\mathrm{S}=$$\mathrm{secA}(1-\mathrm{sinA})(\mathrm{secA}+\mathrm{tanA})$

$=$$\frac{(1-\mathrm{sinA})}{\mathrm{cosA}}\left(\frac{1}{\mathrm{cosA}}+\frac{\mathrm{sinA}}{\mathrm{cosA}}\right)$ $\left(\because \mathrm{secA}=\frac{1}{\mathrm{cosA}}\right)$ and $\left(\mathrm{tanA}=\frac{\mathrm{sinA}}{\mathrm{cosA}}\right)$

$=$$\frac{(1-\mathrm{sinA})}{\mathrm{cosA}}\left(\frac{1+\mathrm{sinA}}{\mathrm{cosA}}\right)$

$=$$\frac{1-{\sin }^2\mathrm{A}}{{\cos }^2\mathrm{A}}$ $\left[\because \left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)={\mathrm{a}}^2-{\mathrm{b}}^2\right]$

$=$$\frac{{\cos }^2\mathrm{A}}{{\cos }^2\mathrm{A}}$ $\left(\because 1-{\sin }^2\mathrm{A}={\cos }^2\mathrm{A}\right)$

$=$$1$

$\therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}$

Thus,$\mathrm{secA}(1-\mathrm{sinA})(\mathrm{secA}+\mathrm{tanA})=1$

Hence Proved