# Question 14663

Aug 4, 2016

The Proof is given below in Explanation.

#### Explanation:

We will use the Identity $: {\sec}^{2} A - {\tan}^{2} A = 1$.

Let us divide the $N r .$ and $D r .$ of L.H.S by $\cos A$ to get,

The L.H.S. ={(sinA+1-cosA)/cosA}/{(sinA-1+cosA)/cosA

$= \frac{\sin \frac{A}{\cos} A + \frac{1}{\cos} A - \cos \frac{A}{\cos} A}{\sin \frac{A}{\cos} A - \frac{1}{\cos} A + \cos \frac{A}{\cos} A}$

$= \frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}$

$= \frac{\tan A + \sec A - \left({\sec}^{2} A - {\tan}^{2} A\right)}{\tan A - \sec A + 1}$

$= \frac{\left(\tan A + \sec A\right) - \left(\sec A + \tan A\right) \left(\sec A - \tan A\right)}{\tan A - \sec A + 1}$

={(secA+tanA)cancel((1-secA+tanA))}/cancel((tanA-secA+1)#

$= \sec A + \tan A$

$=$ The R.H.S.

Hence, the Proof. Enjoy Maths.!

$I {I}^{n d}$ Method :-

We have, $1 - {\sin}^{2} A = {\cos}^{2} A$

$\Rightarrow \left(1 + \sin A\right) \left(1 - \sin A\right) = {\cos}^{2} A$

$\Rightarrow \frac{1 + \sin A}{\cos} A = \cos \frac{A}{1 - \sin A}$

$\therefore$ Each Rratio $= \frac{\left(1 + \sin A\right) - \left(\cos A\right)}{\left(\cos A\right) - \left(1 - \sin A\right)}$, i.e.,

Each Ratio$= \frac{1 + \sin A - \cos A}{\cos A - 1 + \sin A}$.

In particular, $\frac{1 + \sin A}{\cos} A = \frac{1 + \sin A - \cos A}{\cos A - 1 + \sin A}$, or,

$\frac{1}{\cos} A + \sin \frac{A}{\cos} A = \frac{1 + \sin A - \cos A}{\cos A - 1 + \sin A}$.

$\Rightarrow \sec A + \tan A = \frac{1 + \sin A - \cos A}{\cos A - 1 + \sin A}$.