# Prove that `sec^4(x)-tan^4(x)=1+tan^2(x)` .

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### 2 Answers

The identity `sec^4 x - tan^4 x = 1 + tan^2 x` has to be proved.

`sec^4 x - tan^4 x`

=> `(sec^2x - tan^2x)(sec^2x + tan^2x)`

=> `(1/(cos^2x) - (sin^2x)/(cos^2x))(1/(cos^2x) + (sin^2x)/(cos^2x))`

=> `((1 - sin^2x)/(cos^2x))((1 + sin^2x)/(cos^2x))`

=> `((cos^2x)/(cos^2x))((1 + sin^2x)/(cos^2x))`

=> `(1 + sin^2x)/(cos^2x)`

=> `sec^2x + tan^2x`

=> `1 + tan^2x + tan^2x`

=> `1 + 2*tan^2x`

**It is seen that `sec^4 x - tan^4 x = 1 + tan^2 x` is not an identity, instead: `sec^4x - tan^4x = 1 + 2*tan^2x` **

We know sec^2x = 1+tan^2x and (x^2-y^2)=(x-y)(x+y)

sec^4x-tan^4x

=(sec^2x-tan^2x)(sec^2x+tan^2x)

=(1+tan^2x-tan2^x)(1+tan^2x+tan^2x)

=(1)(1+2tan^2x)

=** (1+2tan^2x)**

Your question should be modfied as;

sec^4x-tan^4x