SSC CGL 16 Aug 2015 Shift 2 Paper

1 + $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ = 3 sinθ . cosθ
Dividing both sides by $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ
$\frac{1}{\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }}+\frac{\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }}{\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }}$ = $\frac{3\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }}{\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }}$
⇒ $\mathit{c}\mathit{o}\mathit{s}\mathit{e}{\mathit{c}}^2$ θ + $\mathit{c}\mathit{o}{\mathit{t}}^2$ θ = 3 cot θ
⇒ 1 + $\mathit{c}\mathit{o}{\mathit{t}}^2$ θ + $\mathit{c}\mathit{o}{\mathit{t}}^2$ θ = 3 cot θ
⇒ 2 $\mathit{c}\mathit{o}{\mathit{t}}^2$ θ - 3 cot θ + 1 = 0
⇒ 2 $\mathit{c}\mathit{o}{\mathit{t}}^2$ θ - 2 cot θ - cot θ + 1
= 0
⇒ 2 $\mathit{c}\mathit{o}{\mathit{t}}^2$ θ (cot θ - 1) - 1 (cot θ -1) = 0
⇒ (2 cot θ - 1) (cot θ - 1) = 0
⇒ cot θ = $\frac{1}{2}$ or 1