Question

prove that root under 1 + sin theta upon 1 minus sin theta gives tan theta + sec theta​

Answer 1

To Prove :

• $\sqrt{ \frac{1 + \sin( \theta ) }{1 - \sin( \theta ) } } = \tan( \theta ) + \sec( \theta )$

Proof :

We have been given,

L.H.S =

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

But,

We know that,

• $(x + y)(x + y) = {(x + y)}^{2}$
• $(x + y)(x - y) = { x}^{2} - {y}^{2}$

Therefore,

We get,

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

But,

We know that,

• $1 - { \sin }^{2} \alpha = { \cos }^{2} \alpha$

Therefore,

We get,

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

= R.H.S

Thus,

• L.H.S = R.H.S

Hence, $\large\bold{Proved}$