| If \[\alpha \] is an acute angle and \[2\sin \alpha +15co{{s}^{2}}\alpha =7,\] then the value of cot \[\alpha \]is |
A) \[\frac{5}{4}\]
B) \[\frac{4}{3}\]
C) \[\frac{3}{4}\]
D) \[\frac{4}{5}\]
Correct Answer: C
Solution :
| Here, \[2\sin \alpha +15{{\cos }^{2}}\alpha =7\] |
| \[\Rightarrow \] \[2\sin \alpha +15\,(1-{{\sin }^{2}}\alpha )-7=0\] |
| \[\Rightarrow \] \[2\sin \alpha +15-15{{\sin }^{2}}\alpha -7=0\] |
| \[\Rightarrow \] \[15{{\sin }^{2}}\alpha -2\sin \alpha -8=0\] |
| \[\Rightarrow \] \[15{{\sin }^{2}}\alpha +10\sin \alpha -12sin\alpha -8=0\] |
| \[\Rightarrow \] \[5sin\alpha \,(3\sin \alpha +2)-4\,(3\sin \alpha +2)=0\] |
| \[\Rightarrow \] \[(5\sin \alpha -4)(3\sin \alpha +2)=0\]\[\Rightarrow \]\[\sin \alpha =\frac{4}{5}\] |
| \[\therefore \] \[\cot \alpha =\frac{3}{4}\] |
You need to login to perform this action.
You will be redirected in
3 sec
