180
If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$, then prove that $m^2 - n^2 = 4\sqrt{mn}$.
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$LHS = m^2 - n^2$
$= (\tan \theta + \sin \theta)^2 - (\tan \theta - \sin \theta)^2$
$= 4 \tan \theta \sin \theta$ ($1$ mark)
$= 4 \sqrt{\tan^2 \theta \sin^2 \theta}$ ($1/2$ mark)
$= 4 \sqrt{\tan^2 \theta (1 - \cos^2 \theta)}$ ($1/2$ mark)
$= 4 \sqrt{\tan^2 \theta - \sin^2 \theta}$ ($1/2$ mark)
$= 4 \sqrt{(\tan \theta + \sin \theta)(\tan \theta - \sin \theta)}$
$= 4 \sqrt{mn} = RHS$ ($1/2$ mark)
$= (\tan \theta + \sin \theta)^2 - (\tan \theta - \sin \theta)^2$
$= 4 \tan \theta \sin \theta$ ($1$ mark)
$= 4 \sqrt{\tan^2 \theta \sin^2 \theta}$ ($1/2$ mark)
$= 4 \sqrt{\tan^2 \theta (1 - \cos^2 \theta)}$ ($1/2$ mark)
$= 4 \sqrt{\tan^2 \theta - \sin^2 \theta}$ ($1/2$ mark)
$= 4 \sqrt{(\tan \theta + \sin \theta)(\tan \theta - \sin \theta)}$
$= 4 \sqrt{mn} = RHS$ ($1/2$ mark)