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Integration by Parts

Prove that ...

Question

Prove that $\frac{1 + sec A}{sec A} = \frac{{sin}^{2} A}{1 - cos A}$

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Solution

Let the L.H.S.
$\frac{1 + s e c A}{s e c A}$

$= \frac{1 + \frac{1}{c o s A}}{(\frac{1}{c o s A})}$

$= \frac{\frac{(1 + c o s A)}{c o s A}}{(\frac{1}{c o s A})}$

$= 1 + c o s A$

$= \frac{1 + c o s A}{1 - c o s A} (1 - c o s A)$

$= \frac{1 - c o s^{2} x}{1 - c o s x}$

$= \frac{s i n^{2} A}{1 - c o s A}$

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