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$\cos 2\theta\cos 2\phi+\sin^2(\theta-\phi)-\sin^2(\theta+\phi)$ is equal to

$\begin{array}{1 1}(A)\;\sin 2(\theta+\phi)&(B)\;\cos 2(\theta+\phi)\\(C)\;\sin 2(\theta-\phi)&(D)\;\cos 2(\theta-\phi)\end{array} $

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  • $\sin^2A-\sin^2B=\sin(A+B).\sin (A-B)$
$\cos 2\theta\cos 2\phi+\sin^2(\theta-\phi)-\sin^2(\theta+\phi)$
$\cos 2\theta\cos 2\phi+\sin 2\theta.\sin(-2\phi)$
$\cos 2\theta\cos 2\phi-\sin 2\theta.\sin2\phi$
$\cos (2\theta+2\phi)$
$\cos 2(\theta+\phi)$
Hence (B) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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