A) \[0,-\frac{1}{2}\]
B) \[\frac{1}{2},-1\]
C) \[1,0\]
D) \[\frac{1}{2},-\frac{1}{2}\]
Correct Answer: D
Solution :
Since, A and B are complementary angles, then \[A+B=90{}^\circ \] Now, \[\cos A\,\,\cos B\,=\cos A\,\,\cos (90{}^\circ -A)\] \[=\cos A\sin A=\frac{1}{2}\sin 2A\] Since, \[-1\le \sin 2A\le 1\] Hence, \[-\frac{1}{2}\le \frac{1}{2}\sin 2A\le \frac{1}{2}\] Thus, greatest and least values of \[cos\text{ }A\text{ }cos\text{ }B\] are \[\frac{1}{2}\] and \[-\frac{1}{2}\].
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