Question

find the value of √1+sin theta /1-sin theta​

Answer 1

Solution :

$\sqrt{\dfrac{1\: + \: sin\theta}{1\: - \: sin\theta}}$

$\sqrt{\dfrac{1\:+\: sin\theta}{1\: -\: sin\theta}\times\dfrac{1\: +\: sin\theta}{1\: + \: sin\theta}}$

$\sqrt{\dfrac{(1\: + \: sin\theta)^{2}}{1^{2}\: - \: sin^{2}\theta}}$

$\sqrt{\dfrac{(1\: + \: sin\theta)^{2}}{Cos^2\theta}}$

$\dfrac{1\: + \: sin\theta}{Cos\theta}$

$\dfrac{1}{Cos\theta}\: + \: \dfrac{sin\theta}{cos\theta}$

$sec\: \theta \: + \: tan\: \theta$

Answer 2

QUESTION

value of

$\sqrt{ \frac{ {1 + \sin( \theta) } }{ 1 - \sin( \theta) } }$

FORMULAS USED:

→» (a + b) (a - b) = a² - b²

→» (a + b)(a + b) = (a + b)²

→» sin²θ + cos²θ = 1

cos²θ = 1 - sin²θ

$\frac{ \sin( \theta) }{ \cos( \theta) } = \tan( \theta)$

$\frac{1}{ \cos( \theta) } = \sec( \theta)$

ANSWER:

multiply and divide with

=${1 + \sin( \theta) }$

$\sqrt{ \frac{1 + \sin( \theta) }{1 - \sin( \theta )} \times \frac{1 + \sin( \theta) }{1 + \sin( \theta) } }$

$\sqrt{ \frac{( {1 + \sin( \theta) )}^{2} }{ {1}^{2} - { \sin( \theta) }^{2} } }$

By using radical rules

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

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

=$\frac{1 + \sin( \theta) }{ \cos( \theta) }$

$= \frac{1}{ \cos( \theta) } + \frac{ \sin( \theta) }{ \cos( \theta) }$

=$\sec( \theta) + \tan( \theta)$

CONCEPTS USED

→» trigonometric identities

→» trigonometric ratios

→» rationalization