A) \[0\]
B) \[-1\]
C) \[-\frac{1}{2}\]
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
Since \[\sin \theta .\cos \theta =\frac{1}{2}2\sin \theta .\cos \theta =\frac{1}{2}[\sin 2\theta ]\] Minimum value of \[\sin \,2\theta =-1\] \[\therefore \] Minimum value of \[\sin \theta \,\cos \theta =\frac{1}{2}[-1]=-\frac{1}{2}\]You need to login to perform this action.
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