Question
If $\tan\theta+\sin\theta=\text{m}$ and $\tan\theta-\sin\theta=\text{n},$ then prove that $\text{m}^2-\text{n}^2=4\sin\theta\tan\theta$

Answer

Given that: $\tan\theta+\sin\theta=\text{m}$ and $\tan\theta-\sin\theta=\text{n}$
$\text{L.H.S.}=\text{m}^2-\text{n}^2=(\text{m + n})(\text{m}-\text{n})$
$=[(\tan\theta+\sin\theta)+(\tan\theta-\sin\theta)].[(\tan\theta+\sin\theta)-(\tan\theta-\sin\theta)]$
$=(\tan\theta+\sin\theta+\tan\theta-\sin\theta).(\tan\theta+\sin\theta-\tan\theta+\sin\theta)$
$=2\tan\theta.2\sin\theta=4\sin\theta\tan\theta=\text{R.H.S.}$
$\text{L.H.S. = R.H.S.}$ Hence proved.

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