Question

If \[\mathrm{A}+\mathrm{B}=\frac{\pi}{2}\] then the maximum value of cosA.cosB is

पर्याय

• \[\frac{1}{\sqrt{2}}\]

• \[\frac{1}{2}\]

• \[-\frac{1}{2}\]

• \[-\frac{1}{\sqrt{2}}\]

Answer 1

\[\frac{1}{2}\]

Explanation:

Let y = cos A cos B

\[=\cos\mathrm{A}\cos\left(\frac{\pi}{2}-\mathrm{A}\right)\quad\ldots\left[\because\mathrm{A}+\mathrm{B}=\frac{\pi}{2}\right]\]

= cos A sin A

\[=\frac{1}{2}\cdot2\cdot\sin\mathrm{A}\cos\mathrm{A}\]

\[=\frac{1}{2}\sin2\mathrm{A}\]

Since −1≤ sin x ≤ 1

∴ Maximum value of y is \[\frac{1}{2}\]