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If in a $\Delta ABC$ $\angle C=90^{\circ}$ then the maximum value of $\sin A\sin B$ is

$(a)\;\large\frac{1}{2}$$\qquad(b)\;1\qquad(c)\;2\qquad(d)\;None\;of\;these$

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1 Answer

$\sin A\sin B=\large\frac{1}{2}$$\times 2\sin A\sin B$
$\qquad\quad\quad\;=\large\frac{1}{2}$$[\cos(A-B)-\cos(A+B)]$
$\qquad\quad\quad\;=\large\frac{1}{2}$$[\cos(A-B)-\cos 90^{\circ}]$
$\qquad\quad\quad\;=\large\frac{1}{2}$$[\cos(A-B)\leq \large\frac{1}{2}$
$\Rightarrow$ Maximum value of $\sin A\sin B=\large\frac{1}{2}$
Hence (a) is the correct answer.
answered Oct 7, 2013 by sreemathi.v
 

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