Question

SinA - CosA/SinA+CosA + SinA +CosA/SinA -CosA = 2/2 Sin^2A -1​

Answer 1

Answer:

Step-by-step explanation:

$\frac{sinA - cosA}{sinA+cosA} + \frac{sinA+cosA}{sinA-cosA} = \frac{2}{2sinA^{2}-1 }$

Rationalize the denominators

$\frac{(sinA-cosA)(sinA-cosA) + (sinA+cosA)(sinA+cosA)}{(sinA+cosA)(sinA-cosA)}$

$\frac{(sinA-cosA)^2 + (sinA+cosA)^2}{(sinA^2-cosA^2)}$

Expanding the equation

$\frac{(sinA^2+cosA^2-2sinA.cosA) + (sinA^2+cosA^2-2sinA.cosA)}{sinA^2-cosA^2}$

$\frac{2sinA^2+2cosA^2}{sinA^2-cosA^2}$

$\frac{2(sinA^2+cosA^2)}{sinA^2-cosA^2}$                             ($sinA^2 + cosA^2 = 1$)

Therefore,

$\frac{2}{sinA^2-cosA^2}$

now, $cosA^2 = 1- sinA^2$ substituting this in the above equation we get,

$\frac{2}{sinA^2-(1-sinA^2)}$

$\frac{2}{sinA^2-1+sinA^2}$

$\frac{2}{2sinA^2-1}$ =  $\frac{2}{2sinA^2-1}$

hence LHS = RHS.