Question

1+cos2A /1-cos2A=cotA.cotA​

Answer 1

See Explanation

Explanation:

`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. \\ \\ (1 + \cos2A)/(1 - \cos2A) = \cot \: A. \cot \: A \\ LHS = (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.`