Answer:
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Step-by-step explanation:
[tex]formulae \: to \: be \: used : \\ \cos 2 \theta = 2 { \cos}^{2} \theta - 1 \\ and \\ \cos 2 \theta = 1 - 2 { \cos}^{2} \theta \\ \\ now \: let \: us \: prove \: the \: given \: trigonometrical \: \\ identity. \\ \\ \frac{1 + \cos2A}{1 - \cos2A} = \cot \: A. \cot \: A \\ LHS = \frac{1 + \cos2A}{1 - \cos2A} \\ \\ = \frac{1 +2 { \cos}^{2} A- 1 }{1 - (1 - 2 { \sin}^{2} A)} \\ \\ = \frac{1 +2 { \cos}^{2} A- 1 }{1 - 1 + 2 { \sin}^{2} A} \\ \\ = \frac{2 { \cos}^{2} A }{2 { \sin}^{2} A} \\ \\ = \frac{{ \cos}^{2} A }{ { \sin}^{2} A} \\ \\ = { \cot}^{2} A \\ \\ = { \cot} A .{ \cot} A \\ \\ = RHS \\ \\ hence \: proved.[/tex]
