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Prove that :
$(\tan \alpha + \frac{1}{\cos \alpha})^2 + (\tan \alpha - \frac{1}{\cos \alpha})^2 = 2 (\frac{1+ \sin^2 \alpha}{1- \sin^2 \alpha})$
$(\tan \alpha + \frac{1}{\cos \alpha})^2 + (\tan \alpha - \frac{1}{\cos \alpha})^2 = 2 (\frac{1+ \sin^2 \alpha}{1- \sin^2 \alpha})$
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$LHS = (\frac{\sin \alpha}{\cos \alpha} + \frac{1}{\cos \alpha})^2 + (\frac{\sin \alpha}{\cos \alpha} - \frac{1}{\cos \alpha})^2$
$= (\frac{\sin \alpha+1}{\cos \alpha})^2 + (\frac{\sin \alpha-1}{\cos \alpha})^2$
$= \frac{\sin^2\alpha+2 \sin \alpha+1+\sin^2\alpha-2 \sin \alpha+1}{1- \sin^2\alpha}$
$= \frac{2(1+\sin^2\alpha)}{1- \sin^2\alpha} = RHS$
$= (\frac{\sin \alpha+1}{\cos \alpha})^2 + (\frac{\sin \alpha-1}{\cos \alpha})^2$
$= \frac{\sin^2\alpha+2 \sin \alpha+1+\sin^2\alpha-2 \sin \alpha+1}{1- \sin^2\alpha}$
$= \frac{2(1+\sin^2\alpha)}{1- \sin^2\alpha} = RHS$