If p and q are the lengths of the perpendiculars from the origin to the straight lines
$\begin{array}{l}\mathrm{xsec}\mathrm{\alpha }+\mathrm{ycosec}\mathrm{\alpha }=\mathrm{a}\text{ and }\\ \mathrm{xcos}\mathrm{\alpha }-\mathrm{ysin}\mathrm{\alpha }=\mathrm{acos}2\mathrm{\alpha }\end{array}$
prove that $4{\mathrm{p}}^2+{\mathrm{q}}^2={\mathrm{a}}^2$

The perpendicular distance from (0, 0) to the line ax + by + c = 0 is
$\mathrm{d}=\frac{\left|\mathrm{c}\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$
Given st.line $\mathrm{xsec}\mathrm{\alpha }+\mathrm{ycosec}\mathrm{\alpha }=\mathrm{a}$
The perpendicular distance from (0, 0) to the $\mathrm{xsec}\mathrm{\alpha }+\mathrm{ycosec}\mathrm{\alpha }=\mathrm{a} \mathrm{is} ‘\mathrm{p}’$
$\begin{array}{l}\therefore \mathrm{p}=\frac{|-\mathrm{a}|}{\sqrt{{\sec }^2\mathrm{\alpha }+{\mathrm{cosec}}^2\mathrm{\alpha }}}\\ =\frac{\left|\mathrm{a}\right|}{\sqrt{\frac{1}{{\cos }^2\mathrm{\alpha }}}+\frac{1}{{\sin }^2\mathrm{\alpha }}}=\mathrm{acos}\mathrm{\alpha sin}\mathrm{\alpha }\\ \Rightarrow \mathrm{p}=\mathrm{a}\frac{1}{2}(2\sin \mathrm{\alpha cos}\mathrm{\alpha })\Rightarrow 2\mathrm{p}=\mathrm{asin}2\mathrm{\alpha }\\ \Rightarrow 4{\mathrm{p}}^2={\mathrm{a}}^2{\sin }^{2}2\mathrm{\alpha } ....\left(1\right)\end{array}$
$\text{ And the given st.line }\mathrm{xcos}\mathrm{\alpha }-\mathrm{ysin}\mathrm{\alpha }=\mathrm{acos}2\mathrm{\alpha }$
the perpendicular distance from (0, 0) to the
$\mathrm{xcos}\mathrm{\alpha }-\mathrm{ysin}\mathrm{\alpha }=\mathrm{acos}2\mathrm{\alpha } \mathrm{is} ‘\mathrm{q}’$
$\begin{array}{l}\therefore \mathrm{q}=\frac{\mid -\mathrm{acos}2\mathrm{\alpha I}}{\sqrt{{\cos }^2\mathrm{\alpha }+{\sin }^2\mathrm{\alpha }}}\\ \mathrm{q}=\mathrm{acos}2\mathrm{\alpha }\Rightarrow {\mathrm{q}}^2={\mathrm{a}}^2{\cos }^{2}2\mathrm{\alpha }\end{array}$
(1) + (2)
$\begin{array}{l}4{\mathrm{p}}^2+{\mathrm{q}}^2={\mathrm{a}}^2{\sin }^{2}2\mathrm{\alpha }+{\mathrm{a}}^2{\cos }^{2}2\mathrm{\alpha }\\ ={\mathrm{a}}^2\left[{\sin }^{2}2\mathrm{\alpha }+{\cos }^{2}2\mathrm{\alpha }\right]={\mathrm{a}}^2\\ \therefore 4{\mathrm{p}}^2+{\mathrm{q}}^2={\mathrm{a}}^2\end{array}$