Question

Question #3a14d

Answer 1

Given
`cosecx+sinx=2`

`=>1/sinx+sinx=2`

`=>1+sin^2x=2sinx`

`=>1+sin^2x-2sinx=0`

`=>(1-sinx)^2=0`

`=>1-sinx=0`

`=>sinx=1`

and `cosecx=1`

Now

`cosec^nx+sin^nx`

`=1^n+1^n=2`