Question #18ffc

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Answer 1 · 2016-11-27T09:27:29

See proof below

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 · 2016-11-27T11:06:05

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

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

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

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

Proved

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