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If `sin x+cosec x=2` then `sin^n x +cosec^n x=?`

Answer» `sinx+cosecx = 2`
Taking square on both sides,
`sin^2x+cosec^2x+2sinxcosecx = 4`
`=>sin^2x+cosec^2x+2 = 4` (As `sinxcosecx = 1`)
`=>sin^2x+cosec^2x = 2`
Similarly, we can show that,
`sin^4x+cosec^4x = 2`
Now, `sin^3x+cosec^3x = (sinx+cosecx)(sin^2x+cosec^2x-sinxcosecx)`
`=2(2-1) = 2`
`:. sin^nx+cosec^nx = 2`


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