In a triangle $A B C$ , the side $c$ has two values. Prove that the values satisfy the equation
$\frac{{(a + b)}^{2}}{1 + cos C} + \frac{{(b - a)}^{2}}{1 - cos C} = \frac{{2 a}^{2}}{{sin}^{2} A}$

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Solution

$L . H . S . = \frac{{(a + b)}^{2} + {(b - a)}^{2} - cos C .4 a b}{1 - {cos}^{2} C}$
$= \frac{2 [a^{2} + b^{2} - 2 a b cos C]}{{sin}^{2} C}$
$= 2 {(\frac{a}{sin A})}^{2}$

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