Question

Show that $(\mathrm{cosecA}-\mathrm{sinA})(\mathrm{secA}-\mathrm{cosA})=\frac{1}{\mathrm{tanA}+\mathrm{cotA}.}$

Answer 1

To prove:$(\mathrm{cosecA}-\mathrm{sinA})(\mathrm{secA}-\mathrm{cosA})=\frac{1}{\mathrm{tanA}+\mathrm{cotA}}$

$\mathrm{L}.\mathrm{H}.\mathrm{S}=$$(\mathrm{cosecA}-\mathrm{sinA})(\mathrm{secA}-\mathrm{cosA})$

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

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

$=$$\left(\frac{{\cos }^2\mathrm{A}}{\mathrm{sinA}}\right)\left(\frac{{\sin }^2\mathrm{A}}{\mathrm{cosA}}\right)$ (Identity used : ${\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1$)

$=$$\frac{\mathrm{cosA}\times \mathrm{cosA}}{\mathrm{sinA}}\times \frac{\mathrm{sinA}\times \mathrm{sinA}}{\mathrm{cosA}}$

$=$ $\mathrm{cosAsinA}$

$\mathrm{R}.\mathrm{H}.\mathrm{S}=$$\frac{1}{\mathrm{tanA}+\mathrm{cotA}}$

$=\frac{1}{{\displaystyle \frac{\mathrm{sinA}}{\mathrm{cosA}}}+{\displaystyle \frac{\mathrm{cosA}}{\mathrm{sinA}}}}$

$=\frac{1}{{\displaystyle \frac{{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}}{\mathrm{cosAsinA}}}}$

$=\frac{1}{{\displaystyle \frac{1}{\mathrm{cosAsinA}}}}$ (Identity used: ${\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1$)

$=\mathrm{cosAsinA}$

It is evident from above $\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}$

Thus,$(\mathrm{cosecA}-\mathrm{sinA})(\mathrm{secA}-\mathrm{cosA})=\frac{1}{\mathrm{tanA}+\mathrm{cotA}}$. Hence proved