A) \[0\]
B) \[-\frac{1}{2}\]
C) \[1\]
D) \[-1\]
Correct Answer: B
Solution :
Given, \[A+B+C=\pi \] Also, \[\sin \,C+sin\,A\,cos\,B=0\] \[\therefore \]\[\sin \,[\pi -(A+B)]+\sin A\,\cos B=0\] \[\Rightarrow \] \[\sin (A+B)\,+\sin A\,\cos \,B=0\] \[\Rightarrow \]\[\sin A\,\cos B+\cos A\,\sin B+\sin A\,\cos B=0\] \[\Rightarrow \]\[-2\sin A\,\cos B=\cos A\,\sin B\] \[\therefore \]\[\tan A\,\cot B=\frac{\sin A}{\cos A}.\frac{\cos B}{\sin B}\] \[=\frac{\sin A\cos \,B}{-2\sin \,A\,\cos B}=-\frac{1}{2}\]You need to login to perform this action.
You will be redirected in
3 sec