Question

1+sinQ/1-sinQ= (secQ+tanQ) ²​

Answer 1

$\frac{1 + sinQ}{1 - sinQ} \\ \\ = \frac{( {1 + sinQ})^{2} }{1 - {sin}^{2} Q} \\ \\ = \frac{( {1 + sinQ})^{2} }{ {cos}^{2} Q} \\ \\ = ( \frac{1 + sinQ}{cosQ} ) ^{2} \\ \\ = ( {secQ+ tanQ})^{2}$