Question

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

Answer 1

We have `( cosec theta - sin theta ) = a^3`

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

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

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

Again, `(sec theta - cos theta ) = b^3`

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

      =` ((1-cos^2 theta))/ cos theta`

      =` (sin^2 theta)/cos theta`

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

Now , LHS  = `a^2 b^2 (a^2 + b^2 ) `

  =` a^3 (ab^2) + ( a^2 b^2 ) b^3 `

=`a^3 ( ab^2 ) + ( a^2 b^2 ) b^3 `

=`(cos^2 theta)/(sin theta) xx [(cos ^(2/3) theta)/(sin^(1/3) theta) xx (sin ^(4/3)theta)/(cos ^(2/3) theta)] + [ ( cos ^(4/3) theta theta)/(sin ^(2/3) theta)xx(sin^(2/3)theta)/(cos ^(1/3)theta)] xx sin^2 theta/ cos theta`

 =`cos^2 theta / sin theta xx sin theta + cos theta xx sin^2theta / costheta`

 =`cos^2 theta + sin^2 theta = 1`

= RHS
Hence, proved