# Prove that (cosA+sinA)/(cosA-sinA)+(cosA-sinA)/(cosA+sinA)=2sec2A?

Jun 25, 2016

#### Explanation:

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

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

= $\frac{{\cos}^{2} A + 2 \cos A \sin A + {\sin}^{2} A + {\cos}^{2} A - 2 \cos A \sin A + {\sin}^{2} A}{{\cos}^{2} A - {\sin}^{2} A}$

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

= $\frac{2 \times 1}{\cos 2 A} = 2 \sec 2 A$

(note $\cos 2 A = {\cos}^{2} A - {\sin}^{2} A = 2 {\cos}^{2} A - 1 = 1 - 2 {\sin}^{2} A$)

Jun 25, 2016

see explanation

#### Explanation:

This is not an equation to be solved. The 'key word' here is $\textcolor{b l u e}{\text{Prove}}$

.There are 3 options available to us here in attempting to prove this.

(1) Manipulate the left side into the form of the right side.

(2) Manipulate the right side into the form of the left side.

(3) Manipulate both sides until we reach a common point where both sides are the same.

I am going for option (1).

left side $= \textcolor{b l u e}{\frac{\cos A + \sin A}{\cos A - \sin A}} + \textcolor{red}{\frac{\cos A - \sin A}{\cos A + \sin A}} \ldots \ldots . . \left(A\right)$

We now require to express as a single fraction, which means finding a common denominator.

Let me do this in 2 separate parts.

$\textcolor{b l u e}{\frac{\cos A + \sin A}{\cos A - \sin A}} \times \frac{\cos A + \sin A}{\cos A + \sin A}$

=((cosA+sinA)(cosA+sinA))/((cosA-sinA)(cosA+sinA)
$\textcolor{b l u e}{\text{------------------------------------------------------------}}$

and $\textcolor{red}{\frac{\cos A - \sin A}{\cos A + \sin A}} \times \frac{\cos A - \sin A}{\cos A - \sin A}$

=((cosA-sinA)(cosA-sinA))/((cosA-sinA)(cosA+sinA)
$\textcolor{red}{\text{----------------------------------------------------------------}}$

We can now express the 2 parts as a single fraction in (A)

$\frac{{\left(\cos A + \sin A\right)}^{2} + {\left(\cos A - \sin A\right)}^{2}}{\left(\cos A - \sin A\right) \left(\cos A + \sin A\right)}$

Now distribute the numerator/denominator using FOIL.

$= \frac{{\cos}^{2} A + \cancel{2 \sin A \cos A} + {\sin}^{2} A + {\cos}^{2} A - \cancel{2 \sin A \cos A} + {\sin}^{2} A}{{\cos}^{2} A - {\sin}^{2} A}$

Simplify using the following $\textcolor{\mathmr{and} a n \ge}{\text{Reminders}}$

$\textcolor{m a \ge n t a}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\sin}^{2} \theta + {\cos}^{2} \theta = 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{m a \ge n t a}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and $\textcolor{m a \ge n t a}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sec \theta = \frac{1}{\cos} \theta} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$\textcolor{m a \ge n t a}{\text{----------------------------------------------------}}$

Our expression now simplifies to.

$\frac{2 {\cos}^{2} A + 2 {\sin}^{2} A}{\cos 2 A} = \frac{2 \left({\cos}^{2} A + {\sin}^{2} A\right)}{\cos 2 A} = \frac{2}{\cos 2 A}$

$\Rightarrow \frac{2}{\cos 2 A} = 2 \sec 2 A = \text{ right side"" hence proven}$