Question

Question #6ad0a

Answer 1

Given `cscx-sinx=a`

`=>1/sinx-sinx=a`

`=>(1-sin^2x)/sinx=a`

`=>cos^2x/sinx=a`

Again `secx-cosx=b`

`=>1/cosx-cosx=b`

`=>(1-cos^2x)/cosx=b`

`=>sin^2x/cosx=b`

Now `a^2b^2(a^2+b^2+3)`

`=cos^4x/sin^2x xxsin^4x/cos^2x(cos^4x/sin^2x+sin^4x/cos^2x+3)`

`=sin^2x xxcos^2x((cos^6x+sin^6x)/(sin^2xcos^2x)+3)`

`=cos^6x+sin^6x+3sin^2xcos^2x`

`=(sin^2x)^3+(cos^2x)^3+3sin^2xcos^2x(sin^2x+cos^2x)`

`=(sin^2x+cos^2x)^3=1^3=1`

Proved