Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\frac{(1 + s e c A)}{s e c A} = \frac{s i n^{2} A}{(1 - c o s A)}$
[Hint : Simplify L.H.S and R.H.S separately]

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Solution

$\frac{(1 + s e c A)}{s e c A} = \frac{s i n^{2} A}{(1 - c o s A)} L . H . S . = \frac{(1 + s e c A)}{s e c A} = \frac{(1 + \frac{1}{c o s A)}}{\frac{1}{c o s A}} = \frac{\frac{(c o s A + 1)}{c o s A}}{\frac{1}{c o s A}} = c o s A + 1 R . H . S . = \frac{s i n^{2} A}{(1 - c o s A)} = \frac{(1 - c o s^{2} A)}{(1 - c o s A)} = \frac{(1 - c o s A) (1 + c o s A)}{(1 - c o s A)} = c o s A + 1 L . H . S . = R . H . S .$

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