A) \[\frac{1}{16}\]
B) \[\frac{1}{8}\]
C) \[\frac{3}{16}\]
D) \[\frac{3}{4}\]
Correct Answer: A
Solution :
Given, \[\sin x\cos y=\frac{1}{4}\] and \[3\tan x=4\tan y\] \[\therefore \] \[\frac{3\,\sin \,x}{\cos x}=\frac{4\sin y}{\cos y}\] \[\Rightarrow \] \[3\sin x\cos y=4\sin y\cos x\] \[\Rightarrow \] \[3\times \frac{1}{4}=4\sin y\cos x\] \[\Rightarrow \] \[\sin y\cos x=\frac{3}{16}\] \[\therefore \] \[\sin (x-y)=\sin x\cos y-\cos x\sin y\] \[=\frac{1}{4}-\frac{3}{16}\] \[=\frac{1}{16}\]
You need to login to perform this action.
You will be redirected in
3 sec
