Question

prove that cot^2 A + 1 / cot^2 A - 1 = sec 2A​

Answer 1

$lhs = \frac{ {cot}^{2} \alpha + 1}{ {cot}^{2} \alpha - 1 } \\ \\ = \frac{ \frac{ {cos}^{2} \alpha }{ {sin}^{2}a } + 1}{ \frac{ {cos}^{2} \alpha }{ {sin}^{2} \alpha } - 1} \\ \\ = \frac{ \frac{ {cos}^{2} \alpha + {sin}^{2} \alpha }{ {sin }^{2} \alpha } }{ \frac{ {cos}^{2} \alpha - {sin}^{2} \alpha }{ {sin}^{2} \alpha } } \\ \\ = \frac{1}{ {cos}^{2} \alpha - {sin }^{2} \alpha } . \\ \\ ...........( {sin}^{2} \alpha + {cos}^{2} \alpha = 1) \\ \\ = \frac{1}{cos2 \alpha } \\ \\ ...........( {cos}^{2} \alpha - {sin}^{2} \alpha = cos2 \alpha ) \\ \\ = sec2 \alpha = rhs \\ \\ .............( \frac{1}{cos \alpha } = sec \alpha )$

Answer 2

Answer:

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