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

## Expert Answers 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.

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