Question

√1+sinA/√1-sinA=secA+tanA prove thate​

Answer 1

$\sqrt{ \frac{1 + \sin \: A}{1 - \sin \: A} } = \sec \: A + \tan \: A$

$\sqrt{ \frac{1 + \: \sin \: A }{1 - \sin \: A} }$

Rationalisation of denominator

$= \sqrt{ \frac{1 + \sin \: A }{1 - \sin \: A} \times \frac{1 + \sin \: A }{1 + \sin \: A} }$

$= \sqrt{ \frac{(1 + \sin \: A) {}^{2} }{1 {}^{2} - \sin {}^{2}A } }$

$= \sqrt{ \frac{(1 + \sin \: A) {}^{2} }{1 - \sin {}^{2}A} }$

$= \sqrt{ \frac{(1 + \sin \: A) {}^{2} }{ \cos {}^{2}A } }$

$= \sqrt{( \frac{1 + \sin \: A }{ \cos \: A} ) {}^{2} }$

$= \frac{1 + \sin \: A }{ \cos \: A}$

$= \frac{1}{ \cos \: A } + \frac{ \sin \: A }{ \cos \: A }$

$= \sec \: A + \tan \: A$

Hence proved.

Be Brainly!