SSC CGL 26 Oct 2014 Shift 2 Paper

a cos θ + b sin θ = p
a sin θ – b cos θ = q
On squaring and adding,
${\mathit{a}}^2$ $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ + ${\mathit{b}}^2$ $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ + 2 a b sin θ . cos θ + ${\mathit{a}}^2$ $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ + ${\mathit{b}}^2$ $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ – 2 a b sin θ . cos θ = ${\mathit{p}}^2+{\mathit{q}}^2$
⇒ ${\mathit{a}}^2$ $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ + ${\mathit{a}}^2$ $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ + ${\mathit{b}}^2$ $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ + ${\mathit{b}}^2$ $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ = ${\mathit{p}}^2+{\mathit{q}}^2$
⇒ ${\mathit{a}}^2$ ( $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ + $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ ) + ${\mathit{b}}^2$ ( $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ + $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ ) = ${\mathit{p}}^2+{\mathit{q}}^2$
⇒ ${\mathit{a}}^2+{\mathit{b}}^2$ = ${\mathit{p}}^2+{\mathit{q}}^2$