Question

1. Prove that:
1+sin A
1-sin A
= sec A + tan A.​

Answer 1

We're asked to prove that $\sf{ \sqrt{ \dfrac{1 + sinA}{1 - sinA}} = secA + tanA }$

Rationalising the denominator of LHS,

$\implies{ \sf{\sqrt{ \dfrac{1 + sinA}{1 - sinA} \times \dfrac{1 + sinA}{1 + sinA} }}}$

Now multiplying accordingly,

$\implies{\sf{ \sqrt{ \dfrac{1 + {sin}^{2}A}{ 1 - {sin}^{2}A}}}}$

Now let's use this identity :

$\dag \: \: \: \: \: \boxed{ \sf{ {sin}^{2}A + {cos}^{2}A = 1}}$

$\implies{ \sf{ {cos}^{2}A = 1 - {sin}^{2} A}}$

Let's substitute this in place of denominator.

$\implies{ \sf{ \sqrt{ \dfrac{1 + {sin}^{2}A}{ {cos}^{2}A}}}}$

As both numerator and denominator has squares, it can be taken as a whole

$\implies {\sf{ \sqrt{( \dfrac{1 + sinA}{cosA})^{2} }}}$

Here the sqaure and root will get cancelled obviously, so we get,

$\implies{\sf{ \dfrac{1 + sinA}{cosA}}}$

Which can also be written as

$\implies{\sf{ \dfrac{1}{cosA} + \dfrac{sinA}{cosA}}}$

We know that,

$\sf{ \dfrac{1}{cosA} = secA}$

And, ${\sf{\dfrac{sinA}{cosA} = tanA}}$

Therefore,

$\boxed{ \sf{ \frac{1}{cosA} + \frac{sinA}{cosA} = secA + tanA}}$

Hence Proved!