SSC CGL 19 Oct 2014 Shift 2 Paper

sec α + tan α = 2
⇒ $\frac{1}{\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }}+\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha }}{\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }}$ = 2
⇒ $\frac{1+\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha }}{\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }}$ = 2
⇒ 1 + sin α = 2 cos α
⇒ ${\left(1+\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha }\right)}^2$ = 4 $\mathit{c}\mathit{o}{\mathit{s}}^2$ α
⇒ 1 + $\mathit{s}\mathit{i}{\mathit{n}}^2$ α + 2 sin α = 4(1 – $\mathit{s}\mathit{i}{\mathit{n}}^2$ α )
⇒ 1+ $\mathit{s}\mathit{i}{\mathit{n}}^2$ α + 2 sin α = 4 – 4 $\mathit{s}\mathit{i}{\mathit{n}}^2$ α
⇒ 5 $\mathit{s}\mathit{i}{\mathit{n}}^2$ α + 2 sin α + 1 – 4 = 0
⇒ 5 $\mathit{s}\mathit{i}{\mathit{n}}^2$ α + 2 sin α – 3 = 0
⇒ 5 $\mathit{s}\mathit{i}{\mathit{n}}^2$ α + 5 sin α – 3 sin α – 3 = 0
⇒ 5 sin α (sin α +1) – 3 (sin α +1) = 0
⇒ (5 sin α – 3) (sin α +1) = 0
Since a < 90°,
∴ 5 sin α – 3 = 0
⇒ 5 sin α = 3
⇒ sin α = $\frac{3}{5}$ = 0.6