SSC CGL 6 Sep 2016 Paper

5 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ + 4 $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ = $\frac{9}{2}$
⇒ 10 $\mathit{s}\mathit{i}{\mathit{n}}^2$ θ + 8 $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ = 9
On dividing by $\mathit{c}\mathit{o}{\mathit{s}}^2$ θ
$\frac{10\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }}{\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }}+\frac{8\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }}{\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }}$ = $\frac{9}{\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }}$
= $9\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{\theta }$
⇒ 10 $\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }$ + 8 = 9 (1 + $\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }$ )
⇒ 10 $\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }$ + 8 = 9 + 9 $\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }$
⇒ 10 $\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }$ – 9 $\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }$ = 9 – 8
⇒ $\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }$ = 1 ⇒ tan q = ± 1
Since 0 < θ < $\frac{\mathit{\pi }}{2}$ , ∴ tan θ = 1