Question

If tan theta=4/3, show that √1-sin theta /√1+sin theta = 1/3

Answer 1

$given \: \: \tan \theta = \frac{4}{3} - - - (1)\\ \sec \theta = \sqrt{1 + { \tan }^{2} \theta} = \frac{5}{3} - - - (2) \\ \frac{ \sqrt{1 - \sin \theta } }{\sqrt{1 + \sin \theta } } = \sqrt{ \frac{1 - \sin \theta}{1 + \sin \theta} } \\ = \sqrt{ \frac{ {(1 - \sin \theta)}^{2} }{(1 + \sin \theta)(1 - \sin \theta)} } \\ = \sqrt{ \frac{ {((1 - \sin \theta))}^{2} }{ { \cos }^{2} \theta} } = \frac{(1 - \sin \theta)}{ \cos\theta} \\ = \sec \theta - \tan \theta \\ substituting \: \: from \: \: (1) \: \: and \: \: (2) \\ = \frac{5}{3} - \frac{4}{3} = \frac{1}{3}$

Answer 2

Answer:

$Tan \theta = \frac{Perpendicular}{Base}$

We are given that $tan \theta = \frac{4}{3}$

So, On comparing

Perpendicular = 4

Base = 3

To Find Hypotenuse We will use Pythagoras theorem :

$Hypotenuse^2=Perpendicular^2+Base^2$

$Hypotenuse^2=4^2+3^2$

$Hypotenuse^2=16+9$

$Hypotenuse^2=25$

$Hypotenuse=\sqrt{25}$

$Hypotenuse=5$

$Sin \tehta = \frac{Perpendicular}{Hypotenuse}$

So, $Sin \tehta = \frac{4}{5}$

We are supposed to find $\frac{\sqrt{1-sin \theta}}{\sqrt{1+sin \theta}}$

Substitute the value

$\frac{\sqrt{1-\frac{4}{5}}}{\sqrt{1+\frac{4}{5}}}$

$\frac{\sqrt{\frac{1}{5}}}{\sqrt{\frac{9}{5}}}$

$\frac{1}{3}$

Hence  $\frac{\sqrt{1-sin \theta}}{\sqrt{1+sin \theta}}$ = $\frac{1}{3}$