NIMCET 2018 Question Paper with solutions

Now,
sin θ - cos θ = $\sqrt{2}$ cos θ
⇒ ${\left(\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }-\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\right)}^2$ = $2\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }$
⇒ $\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }+\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }$ - 2 sinθ cosθ = $2\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }$
⇒ $\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }$ = $\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }$ + 2sinθcosθ
⇒ $2\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }$ = $\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }+\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }$ + 2 sinθcosθ
⇒ $2\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }$ = ${\left(\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }+\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\right)}^2$
⇒ $\sqrt{2}$ sin θ = sin θ + cos θ
∴ P = Q [taking positive square root]