A) \[\frac{5}{4}\]
B) \[\frac{3}{4}\]
C) \[\frac{1}{4}\]
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
| [b] \[2\sin \alpha +15co{{s}^{2}}\alpha =7\] \[\Rightarrow \] \[2\sin \alpha +15-15{{\sin }^{2}}\alpha =7\] \[\Rightarrow \] \[15{{\sin }^{2}}\alpha -2\sin \alpha -8=0\] \[\Rightarrow \] \[(5sin\alpha -4)(3sin\alpha +2)=2\] \[\Rightarrow \] \[\sin \alpha =\frac{4}{5}\]or\[\frac{-2}{3}\] \[[\therefore 0\le \alpha \le \frac{\pi }{2}\therefore sin\alpha \,\]cannot be negative] \[\therefore \] \[\cos \alpha =\frac{\sqrt{{{5}^{2}}-{{4}^{2}}}}{5}=\frac{3}{5}\] Then, \[\cot \alpha =\frac{5}{4}=\frac{3}{4}\] |
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