If \[\sin \alpha \sec (30{}^\circ +\alpha )=1,\]\[(0<\alpha <60{}^\circ ),\] then the value of \[\sin \alpha +\cos 2\alpha ,\]is [SSC (10+2) 2011] |
A) \[1\]
B) \[\frac{2+\sqrt{3}}{2\sqrt{3}}\]
C) \[0\]
D) \[2\]
Correct Answer: A
Solution :
\[\sin \alpha \cdot sec(30{}^\circ +\alpha )=1\] |
\[\Rightarrow \] \[\frac{\sin \alpha }{\cos \,\,(30{}^\circ +\alpha )}=1\] |
\[\Rightarrow \]\[\frac{\sin \alpha }{\sin \,\,[90{}^\circ -(30{}^\circ +\alpha )]}=1\] |
\[\Rightarrow \] \[\sin \alpha =\sin \,\,(60{}^\circ -\alpha )\] |
\[\Rightarrow \] \[\alpha =60{}^\circ -\alpha \] |
\[\Rightarrow \] \[2\alpha =60{}^\circ \]\[\Rightarrow \]\[\alpha =30{}^\circ \] |
Hence, \[\sin \alpha +\cos 2\alpha =\sin 30{}^\circ +\cos 60{}^\circ \] |
\[=\frac{1}{2}+\frac{1}{2}=1\] |
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