Question

prove that root 1 minus sin theta by 1 + sin theta equals to sec theta minus tan theta

Answer 1

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

$\sqrt{ \frac{1 - sin}{1 + sin} } \times \sqrt{ \frac{1 - sin}{1 - sin} }$

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

We know that,

${sin}^{2} + {cos}^{2} = 1 \\ = > 1 - {sin}^{2} = {cos}^{2}$

Now we have,

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

$= > \frac{1}{cos} - \frac{sin}{cos}$

$= > sec - tan$

Hence,it is proved that

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

Answer 2

$\huge{\mathfrak{\underline{\underline{\red{Answer :-}}}}}$

To Prove :-

$\large{\sqrt{\frac{1 - sin{\theta}}{1 + sin{\theta}}} = sec{\theta} - tan{\theta}}$

Solution :-

$\large{\sqrt{\frac{1 - sin{\theta}}{1 + sin{\theta}}}}$

✯ Rationalizing the denominator

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

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

Using Formula

$\huge{\boxed{{\gray{{sin}^{2}{\theta} + {cos}^{2}{\theta} =1}}}}$

Now we have,

$\large{\sqrt{ \frac{( {1 - sin}{\theta})^{2} }{ {cos}^{2}{\theta} } } = \: \frac{1 - sin{\theta}}{cos{\theta}}}$

$\large{\frac{1}{cos{\theta}} - \frac{sin{\theta}}{cos{\theta}}}$

Hence,

$\large{\sqrt{\frac{1 - sin{\theta}}{1 + sin{\theta}}} = sec{\theta} - tan{\theta}}$

L.H.S = R.H.S

$\huge{\bf{\boxed{Hence \: \: proved}}}$