Question

Question #18ffc

Answer 1

See proof below

Explanation:

We use `cos^2x+sin^2x=1`

`sin^2x-1=-cos^2x`

`sin^2x=1-cos^2x`

LHS

`=tan^4x-sec^4x=sin^4x/cos^4x-1/cos^4x`

`=(sin^4x-1)/cos^4x=((sin^2x+1)(sin^2x-1))/cos^4x`

`=-(cos^x(sin^2x+1))/(cos^4x)=-(sin^2x+1)/cos^2x`

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

`=1-2/cos^2x=1-2sec^2x = RHS`

Answer 2

`LHS=tan^4x-sec^4x`

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

`=(-1)(sec^2x-1+sec^2x)`

`=1-2sec^2x =RHS`

Proved

Using the identity `sec^2x-tan^2x=1`