NIMCET 2012 Question Paper with solutions

Given, sin (π cos θ) = cos (π sin θ)
⇒ cos (π sin θ) = cos $\left\{\frac{\mathit{\pi }}{2}-\left(\mathit{\pi }\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\right)\right\}$
⇒ π sin θ = ± $\left[\frac{\mathit{\pi }}{2}-\mathit{\pi }\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\right]$
⇒ sin θ + cos θ = $\frac{1}{2}$ (taking + ve sign)
⇒ ${\left(\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }+\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\right)}^2$ = ${\left(\frac{1}{2}\right)}^2$
⇒ $\left(\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }+\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }\right)$ + 2 sin θ . cos θ = $\frac{1}{4}$
⇒ 1 + sin 2θ = $\frac{1}{4}$
⇒ sin 2 θ = $-\frac{3}{4}$ ... (i)
⇒ sin θ = - $\frac{1}{2}$ + cos θ (taking - ve sign)
⇒ cos θ - sin θ = $\frac{1}{2}$
On squaring both sides,
${\left(\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }-\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\right)}^2$ = ${\left(\frac{1}{2}\right)}^2$
⇒ $\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }+\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }$ - 2 sin θ . cos θ = $\frac{1}{4}$
⇒ 1 - sin 2θ = $\frac{1}{4}$
⇒ sin 2θ = $\frac{3}{4}$ ... (ii)
∴ From Eqs. (i) and (ii), we get
sin 2θ = ± $\frac{3}{4}$