(B) Given $A+B=\frac{\pi}{2}$,so $B=\frac{\pi}{2}-A$. Substituting this into the expression,we get $\cos A \cdot \cos B = \cos A \cdot \cos(\frac{\pi}

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(B) Given $A+B=\frac{\pi}{2}$,so $B=\frac{\pi}{2}-A$. Substituting this into the expression,we get $\cos A \cdot \cos B = \cos A \cdot \cos(\frac{\pi}{2}-A)$. Since $\cos(\frac{\pi}{2}-A) = \sin A$,the expression becomes $\cos A \cdot \sin A$. Multiplying and dividing by $2$,we get $\frac{2 \sin A \cos A}{2} = \frac{\sin(2A)}{2}$. The maximum value of $\sin(2A)$ is $1$. Therefore,the maximum value of $\frac{\sin(2A)}{2}$ is $\frac{1}{2}$.

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities

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If $A+B=\frac{\pi}{2}$,then the maximum value of $\cos A \cdot \cos B$ is

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A
$\frac{1}{\sqrt{2}}$
B
$\frac{1}{2}$
C
$-\frac{1}{2}$
D
$-\frac{1}{\sqrt{2}}$

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Class 11

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Trigonometrical Ratios, Functions and Identities

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Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities

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