if `cosec theta - sin theta = a^3`, `sec theta - cos theta = b^3` prove that `a^2 b^2 (a^2 + b^2) = 1`

#### Solution

Given that,

`cosec theta - sin theta = a^3` .....(1)

`sec theta - cos theta = b^3` ......(2)

We have to prove `a^2b^2(a^2 + b^2) = 1`

We know that `sin^2 theta + cos^2 theta = 1`

Now from the first equation, we have

`cosec theta - sin theta = a^3`

`=> 1/sin theta - sin theta = a^3`

`=> (1 - sin^2 theta)/sin theta = a^3`

`=> cos^2 theta/sin theta = a^3`

`=> a = (cos^(2/3) theta)/(sin^(1/3) theta)`

Again from the second equation, we have

`sec theta - cos theta =- b^3`

`=> 1/cos theta - cos theta = b^3`

`=> (1 - cos^2 theta)/cos theta = b^3`

`=> sin^2 theta/cos theta = b^3`

`=> b = (sin^(2/3) theta)/(cos^(1/3) theta)`

Therefore, we have

`a^2b^2 (a^2 + b^2) = (cos^(4/3) theta)/(sin^(2/3) theta cos^(2/3) theta) ((cos^(4/3) theta)/(sin^(2/3) theta) + (sin^(4/3) theta)/(cos^(2/3) theta))`

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

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

`= cos^2 theta + sin^2 theta`

= 1

Hence proved.