A) \[\frac{1}{2}+\frac{1}{\sqrt{3}}\]
B) \[\sqrt{3}+2\]
C) \[\frac{\sqrt{3}}{2}\]
D) \[0\]
Correct Answer: C
Solution :
| [c] We have, \[\cos \alpha =\sqrt{3}/2\,\,\Rightarrow \alpha =30{}^\circ \] \[[\cos 30{}^\circ =\sqrt{3}/2]\] |
| and \[\tan \beta =\frac{1}{\sqrt{3}}\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\beta =30{}^\circ \] \[\left[ \tan 30{}^\circ =\frac{1}{\sqrt{3}} \right]\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\alpha +\beta =30{}^\circ +30{}^\circ =60{}^\circ \] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\sin (\alpha +\beta )=\sin 60{}^\circ =\frac{\sqrt{3}}{2}\] |
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