Question

If
$\tan\alpha = \frac{3}{4}$
Then prove that
$\sqrt{ \frac{1 - \sin\alpha }{1 + \sin\alpha } } = \frac{1}{2}$
Please solve this as soon as possible......​

Answer 1

Step-by-step explanation:

$given \: \\tan \alpha = \frac{3}{4} \\ then \: sin \alpha \: will \: be \: \frac{3}{5} \\ to \: prove: \sqrt{ \frac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) } } = \frac{1}{2} \\ lhs : \\ = \sqrt{ \frac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) } } \\ = \sqrt{ \frac{1 - \frac{3}{5} }{1 + \frac{3}{5} } } \\ = \sqrt{ \frac{ \frac{5 - 3}{5} }{ \frac{5 + 3}{5} } } \\ = \sqrt{ \frac{ \frac{2}{5} }{ \frac{8}{5} } } \\ = \sqrt{ \frac{2}{5} \times \frac{5}{8} } \\ = \sqrt{ \frac{1}{4} } \\ = \frac{ \sqrt{1} }{ \sqrt{4} } \\ = \frac{1}{2} \\ therefore \: lhs = rhs \\ hence \: proved$