Question

prove that 1+sinA/1-sinA=2(secA+tanA)​

Answer 1

Answer:

$\frac{1 + sin \: \alpha }{1 - sin \: \alpha } \times \frac{1 + sin \: \alpha }{1 + sin \: \alpha }$

$\frac{1 + {sin}^{2} \alpha + 2sin \alpha \: \alpha }{1 - {sin}^{2} \alpha }$

$\frac{1 + {sin}^{2} \alpha + 2sin \: \alpha }{ {cos}^{2} \alpha }$

${ \sec }^{2} \alpha + {tan}^{2} \alpha + 2 \tan \alpha \sec \alpha$

$( \sec \alpha + \tan \alpha ) {}^{2}$

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Hope it helps you

Answer 2

$\frac{1 + sinA}{1 - sinA} \\ \\ = \frac{( {1 + sinA})^{2} }{1 - {sin}^{2}A } \\ \\ = \frac{( {1 + sinA})^{2} }{ {cos}^{2} A} \\ \\ = ( \frac{1 + sinA}{cosA} )^{2} \\ \\ = ( {secA + tanA})^{2}$