SSC CGL 16 Aug 2015 Shift 2 Paper

sec θ + tan θ = 2 + 5
Since $\mathit{s}\mathit{e}{\mathit{c}}^2$ θ – $\mathit{t}\mathit{a}{\mathit{n}}^2$ θ = 1
⇒ (sec θ + tan θ ) (sec θ – tan θ ) = 1
⇒ sec θ – tan θ = $\frac{1}{\sqrt{5}+2}$
= $\frac{1}{\sqrt{5}+2}\times \frac{\sqrt{5}-2}{\sqrt{5}-2}$ = $\frac{\sqrt{5}-2}{5-4}$
= $\sqrt{5}$ - 2
∴ sec θ + tan θ + sec θ - tan θ = 2 + $\sqrt{5}$ + $\sqrt{5}$ - 2
⇒ 2 sec θ = $2\sqrt{5}$
⇒ sec θ = $\sqrt{5}$ ... (i)

sec θ + tan θ - (sec θ - tan θ)
= 2 + $\sqrt{5}-\sqrt{5}$ + 2
⇒ 2 tan θ = 4
⇒ tan θ = 2 ... (ii)
∴ sin θ = $\frac{\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}{\mathit{s}\mathit{e}\mathit{c}\mathit{\theta }}$ = $\frac{2}{\sqrt{5}}$