Prove ((sinx+cosx)/(sinx-cosx))+((sinx-cosx)/(sinx+cosx)) = (2sec^2 x/tan^2 x)-1

We'll manage the left side of expression. We notice that
we have to add two fractions whose denominators are not equal. Therefore, we'll multiply
the 1st fraction by the denominator of the second and we'll multiply the 2nd fraction by
the denominator of the first fraction, such as:

[(sin x +
cos x)(sin x + cos x) + (sin x - cos x)(sin x - cos x)]/(sin x + cos x)(sin x - cos
x)

The product from denominator returns the difference of
two squares:

[(sin x + cos x)^2 + (sin x - cos x)^2]/[(sin
x)^2 - (cos x)^2]

We'll expand the square from
numerator:

[(sin x)^2 + 2sin x*cos x + (cos x)^2 + (sin
x)^2 - 2sin x*cos x + (cos x)^2]/[(sin x)^2 - (cos
x)^2]

But, from Pythagorean identity, we'll
get:

(sin x)^2 +  (cos x)^2 =
1

Eliminating like terms from numerator, we'll
get:

2/[(sin x)^2 - (cos
x)^2](1)

We'll manage the right
side:

2(sec x)^2/(tan x)^2 -
1

We know that (sec x)^2 = 1/(cos x)^2 and (tan x)^2 = (sin
x)^2/(cos x)^2

2(sec x)^2/(tan x)^2 = [2/(cos x)^2]/[(sin
x)^2/(cos x)^2]

We'll simplify and we'll
get:

2(sec x)^2/(tan x)^2 - 1 = [2-(sin x)^2]/(sin x)^2
(2)

We notice that the left side (1) is not
equal to the right side (2), therefore the given expression does not represent an
identity.