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

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justaguide eNotes educator| Certified Educator

We have to prove that :((sin x + cos x)/(sin x - cos x)) + ((sin x  - cosx)/(sin x + cos x)) = (2*sec^2x/tan^2x) - 1

((sin x + cos x)/(sin x - cos x)) + ((sin x  - cosx)/(sin x + cos x))

=> ((sin x + cos x)^2 + (sin x - cos x)^2)/(sin x + cos x)(sin x - cos x)

=> ((sin x)^2 + (cos x)^2 + 2*(sin x)(cos x) + (sin x)^2 + (cos x)^2 - 2*(cos x)(sin x))/(sin x + cos x)(sin x - cos x)

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

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

=> (2*sec^2x - tan^2x)/tan^2x

=> [(2/(cos x)^2 - (sin x)^2/(cos x)^2)]/[(sin x)^2/(cos x)^2]

=> [2 - (sin x)^2]/(sin x)^2 ...(2)

As we can see (1) is not equal to (2)

The given relation is not an identity.

giorgiana1976 | Student

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.