A) \[0{}^\circ \]
B) \[45{}^\circ \]
C) \[60{}^\circ \]
D) \[90{}^\circ \]
Correct Answer: B
Solution :
Since \[\tan \alpha =\frac{1}{7}\] \[\therefore \] \[\sin \alpha =\frac{1}{5\sqrt{2}}\] and \[\cos \alpha =\frac{7}{5\sqrt{2}}\] Since \[\sin \beta =\frac{1}{\sqrt{10}}\] \[\therefore \]\[\cos \beta =\frac{3}{\sqrt{10}}\] \[\therefore \]\[\sin (\alpha +2\beta )=\sin \alpha \cos 2\beta +\cos \alpha \sin 2\beta \] \[=\sin \alpha (2{{\cos }^{2}}\beta -1)+\cos \alpha .2\sin \beta \cos \beta \] \[=\frac{1}{5\sqrt{2}}\left( 2\times \frac{9}{10}-1 \right)+\frac{7}{5\sqrt{2}}\times 2\times \frac{1}{\sqrt{10}}\times \frac{3}{\sqrt{10}}\] \[=\frac{1}{5\sqrt{2}}\times \frac{4}{5}+\frac{21}{25\sqrt{2}}\] \[=\frac{4}{25\sqrt{2}}+\frac{21}{25\sqrt{2}}\] \[=\frac{25}{25\sqrt{2}}=\frac{1}{\sqrt{2}}\] \[=\sin {{45}^{o}}\] or \[(\alpha +2\beta )={{45}^{o}}\]
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