SSC CGL 16 Aug 2015 Shift 2 Paper

Expression = 3($\mathit{s}\mathit{i}{\mathit{n}}^4$ θ + $\mathit{c}\mathit{o}{\mathit{s}}^4$ θ) + 2($\mathit{s}\mathit{i}{\mathit{n}}^6$ θ + $\mathit{c}\mathit{o}{\mathit{s}}^6$ θ) + 12 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ
= 3{($\mathit{s}\mathit{i}{\mathit{n}}^2$ θ + $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ )2 – 2 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ . $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ } + 2{($\mathit{s}\mathit{i}{\mathit{n}}^2$ θ + $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ )3 – 3 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ . $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ ($\mathit{s}\mathit{i}{\mathit{n}}^2$ θ + $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ ) + 12 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ . $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ
[Since ${\mathit{a}}^2+{\mathit{b}}^2$ = ${\left(\mathit{a}+\mathit{b}\right)}^2$ – 2ab ; ${\mathit{a}}^3+{\mathit{b}}^3$ = ${\left(\mathit{a}+\mathit{b}\right)}^3$ – 3ab (a + b)]
= 3 (1 – 2 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ . $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ ) + 2 (1 – 3 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ . $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ ) + 12 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ . $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ = 3 – 6 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ . $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ + 2 – 6 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ + 12 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ . $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ = 5