Prove that `sec^4(x)-tan^4(x)=1+tan^2(x)` .
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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`
jeew-m | College Teacher | (Level 1) Educator Emeritus
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