If $p_1,p_2,p_3$ are respectively the perpendiculars from the vertices of a triangle to the opposite sides, then $\frac{\cos A}{p_1}+\frac{\cos B}{p_2}+\frac{\cos C}{p_3}$ is equal to

We have,

$\begin{array}{l}\frac{\cos A}{p_1}+\frac{\cos B}{p_2}+\frac{\cos C}{p_3}\\ =\frac{1}{2\mathrm{\Delta }}(a\cos A+b\cos B+i\cos C)\end{array}$

$\begin{array}{l}=\frac{R}{\mathrm{\Delta }}(\sin A\cos A+\sin B\cos B+\sin C\cos C)\\ =\frac{R}{2\mathrm{\Delta }}(\sin 2A+\sin 2B+\sin 2C)\\ =R\frac{4\sin A\sin B\sin C}{2\mathrm{\Delta }}=\frac{2R\sin A\sin B\sin C}{\mathrm{\Delta }}\\ =\frac{2R}{\mathrm{\Delta }}\times \frac{2\mathrm{\Delta }}{bc}\times \frac{2\mathrm{\Delta }}{ca}\times \frac{2\mathrm{\Delta }}{ab}=\frac{16R{\mathrm{\Delta }}^2}{a^2{b}^2{c}^2}=\frac{16R{\mathrm{\Delta }}^2}{(4R\mathrm{\Delta }{)}^2}=\frac{1}{R}\end{array}$