Find the area of the triangle formed by the straight line $\mathrm{xcos}\mathrm{\alpha }+\mathrm{ysin}\mathrm{\alpha }=\mathrm{p}$ and the coordinate axes.

Given st.line $\mathrm{xcos}\mathrm{x}+\mathrm{ysin}\mathrm{\alpha }=\mathrm{p}$
The area of the triangle formed by line
$\mathrm{ax}+\mathrm{by}+\mathrm{c}=0 \mathrm{with} \mathrm{co}-\mathrm{ordinate} \mathrm{axes} \mathrm{is} \frac{{\mathrm{c}}^2}{2\left|\mathrm{ab}\right|}$
$=\frac{(-\mathrm{p}{)}^2}{2|\cos \mathrm{\alpha sin}\mathrm{\alpha }|}=\frac{{\mathrm{p}}^2}{|\sin 2\mathrm{\alpha }|}={\mathrm{p}}^2|\mathrm{cosec}2\mathrm{\alpha }|$