Question

Prove that 1+Sin theta /1-Sin theta =(sec theta+tan theta) ^ 2​

Answer 1

Answer:

LHS

$\frac{1 + \sin( \alpha ) }{1 - \sin( \alpha ) } = \frac{1 + \sin( \alpha ) }{1 - \sin( \alpha ) } \times \frac{1 + \sin( \alpha ) }{1 + \sin( \alpha ) } =$

$\frac{(1 + sin \: \alpha ) {}^{2} }{1 - sin {}^{2} \alpha } = \frac{(1 + sin \: \alpha ) {}^{2} }{cos {}^{2} \alpha } = ( \frac{1 + sin \: \alpha }{cos \: \alpha } ) {}^{2}$

$= ( \frac{1}{cos \: \alpha } + \frac{sin \: \alpha }{cos \: \alpha } ) {}^{2} = (sec {}^{2} \alpha + tan {}^{2} \alpha ) {}^{2}$

RHS. Proved.