SSC CGL 9 Aug 2015 Shift 1 Paper

x cos θ – sin θ = 1
⇒ x cos θ = 1 + sin θ
⇒ x = $\frac{1}{\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }}+\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}{\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }}$
⇒ x = sex θ + tan θ ... (i)
Since $\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{\theta }-\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }$ = 1
⇒ (sec θ + tan θ) (sec θ - tan θ) = 1
⇒ sec θ - tan θ = $\frac{1}{\mathit{x}}$ ... (ii)
From equation (i) + (ii),
2 sec θ = x + $\frac{1}{\mathit{x}}$ = $\frac{{\mathit{x}}^2+1}{\mathit{x}}$
⇒ sec θ = $\frac{{\mathit{x}}^2+1}{2\mathit{x}}$
From equation (i) – (ii),
2 tan θ = x - $\frac{1}{\mathit{x}}$ = $\frac{{\mathit{x}}^2+1}{\mathit{x}}$
∴ tan θ = $\frac{{\mathit{x}}^2-1}{2\mathit{x}}$
∴ sin θ = $\frac{\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }}{\mathit{s}\mathit{e}\mathit{c}\mathit{\theta }}$
= $\frac{{\mathit{x}}^2+1}{2\mathit{x}}$ × $\frac{2\mathit{x}}{{\mathit{x}}^2+1}$ = $\frac{{\mathit{x}}^2+1}{{\mathit{x}}^2+1}$
∴ Expression = ${\mathit{x}}^2-\left(1+{\mathit{x}}^2\right)$ sin θ
= ${\mathit{x}}^2-\left(1+{\mathit{x}}^2\right)$ × $\frac{{\mathit{x}}^2-1}{{\mathit{x}}^2+1}$
= ${\mathit{x}}^2-{\mathit{x}}^2$ + 1 = 1