Question

Prove that -

√1+sin A/1-sin A = sec A + tan A

Answer 1

Question :-

• $\displaystyle{\implies\sf\sqrt{ \frac{ 1 + sin \: A}{1 - sin \: A }}= sec \: A + tan \: A}$

Answer :-

• $\sf{ L.H.S = R.H.S}$

step by step explanation :-

• ${ \sf{Refer \: in \: the \: attachment !}}$

Answer 2

QUESTION :- Prove that $\sqrt{ \frac{1 + \sin \: A}{1 - \sin \: A}} = \sec \: A + \tan \: A$

SOLUTION :-

$→\sqrt{ \frac{1 + \sin \: A}{1 - \sin \: A}}$

⇒Rationalizing the terms,

$→\sqrt{ \frac{1 + \sin \: A}{1 - \sin \: A} \times \frac{1 + \sin \: A}{1 + \sin \: A}}$

$→ \sqrt{ \frac{{(1 + \sin \: A)}^{2}}{ {1}^{2} + {(\sin \: A)}^{2} } }$

⇒Canceling root for numerator,

$→ \frac{1 + \sin \: A}{ \sqrt{1 - {\sin}^{2} \: A} }$

⇒Canceling root and 2 power of denominator,

$→ \frac{1 + \sin \: A}{\sqrt{ {\cos}^{ \cancel2}\: A} }$

$→ \frac{1 + \sin \: A}{\cos \: A}$

$→ \frac{1}{\cos \: A} + \frac{\sin \: A}{\cos \: A}$

⇒We know that,

$\frac{1}{ \cos \theta } = \sec \theta$ and $\frac{\sin \theta}{ \cos \theta} = \tan \theta$

⇒Therefore,

$= \sec \: A + \tan \: A$

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

LHS = RHS

Hence Proved.