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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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The value of $\cos(45^{\circ}-A)\cos(45^{\circ}-B)-\sin(45^{\circ}-A)\sin(45^{\circ}-B)$ is equal to

$\begin{array}{1 1}(a)\;\sin(A+B)&(b)\;\cos(A+B)\\(c)\;\sin(A-B)&(d)\;\cos(A-B)\end{array}$

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1 Answer

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$\cos(45^{\circ}-A)\cos(45^{\circ}-B)-\sin(45^{\circ}-A)\sin(45^{\circ}-B)$
$\Rightarrow \cos\{(45^{\circ}-A)+(45^{\circ}-B)\}$
$\Rightarrow \cos\{90^{\circ}-(A+B)\}$
$\Rightarrow \sin(A+B)$
Hence (a) is the correct answer.
answered Oct 4, 2013 by sreemathi.v
 

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