NIMCET 2012 Question Paper with solutions

Given, cos (α + β) = $\frac{4}{5}$
and sin (α - β) = $\frac{5}{13}$ where , 0 < α , β < $\frac{\mathit{\pi }}{4}$
Using the identity $\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }+\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }$ = 1
Now , sin (α + β) = $\sqrt{1\mathit{c}\mathit{o}{\mathit{s}}^2\left(\mathit{\alpha }\mathit{\beta }\right)}$ = $\sqrt{1-\frac{16}{25}}$ = $\sqrt{\frac{9}{25}}$
∴ sin (α+ β) = $\frac{3}{5}$
and cos (α - β) = $\sqrt{1-\mathit{s}\mathit{i}{\mathit{n}}^2\left(\mathit{\alpha }-\mathit{\beta }\right)}$
= $\sqrt{1-\frac{25}{169}}$ = $\sqrt{\frac{144}{168}}$
∴ cos (α - β) = $\frac{12}{13}$
Now, tan 2α = tan }(α + β) + (α- β)}
= $\frac{\mathit{t}\mathit{a}\mathit{n}\left(\mathit{\alpha }+\mathit{\beta }\right)+\mathit{t}\mathit{a}\mathit{n}\left(\mathit{\alpha }-\mathit{\beta }\right)}{1-\mathit{t}\mathit{a}\mathit{n}\left(\mathit{\alpha }+\mathit{\beta }\right).\mathit{t}\mathit{a}\mathit{n}\left(\mathit{\alpha }-\mathit{\beta }\right)}$
= $\frac{\frac{\mathit{s}\mathit{i}\mathit{n}\left(\mathit{\alpha }\mathit{\beta }\right)}{\mathit{c}\mathit{o}\mathit{s}\left(\mathit{\alpha }\mathit{\beta }\right)}+\frac{\mathit{s}\mathit{i}\mathit{n}\left(\mathit{\alpha }\mathit{\beta }\right)}{\mathit{c}\mathit{o}\mathit{s}\left(\mathit{\alpha }\mathit{\beta }\right)}}{1-\frac{\mathit{s}\mathit{i}\mathit{n}\left(\mathit{\alpha }+\mathit{\beta }\right)}{\mathit{c}\mathit{o}\mathit{s}\left(\mathit{\alpha }\mathit{\beta }\right)}.\frac{\mathit{s}\mathit{i}\mathit{n}\left(-\mathit{\beta }\right)}{\mathit{c}\mathit{o}\mathit{s}\left(\mathit{\alpha }\mathit{\beta }\right)}}$
= $\frac{\frac{3}{5}\times \frac{5}{4}+\frac{5}{13}\times \frac{13}{12}}{1-\left(\frac{3}{5}\times \frac{5}{4}\right)\left(\frac{5}{13}\times \frac{13}{12}\right)}$
= $\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{15}{4.12}}$
= $\frac{\left(9+5\right)}{12\left(1-\frac{15}{4\times 12}\right)}$
= $\frac{14}{12-\frac{15}{4}}$
= $\frac{14\times 4}{33}$ = $\frac{56}{33}$