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If $\tan \theta+\sin \theta=m$ and $\tan\theta-\sin \theta=n$ then prove that $m^2-n^2=4\sin \theta\tan\theta$

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$\tan\theta+\sin \theta=m$--------(1)
$\tan\theta-\sin \theta=n$-------(2)
After adding (1) & (2) we get
$2\tan \theta=m+n$-----(3)
After subtracting (1) & (2) we get
$2\sin \theta=m-n$-----(4)
After multiplying (3) & (4) we get
$2\tan\theta\times 2\sin \theta=(m+n)(m-n)$
$4\sin \theta\tan\theta=m^2-n^2$
Hence proved
answered Apr 18, 2014 by sreemathi.v

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