If `sinA+cosA=m` and `sin^3 A +cos^3 A = n` then

Options

A. `m^(3) -3m + n=0`
B. ` n^(3) -3n +2m=0`
C. `m^(3) -3m + 2n=0`
D. `m^(3) + 3m + 2n =0`

Answer

C

Explanation

Given sin A + cos A =m
On cubing both sides, we get
`sin^(3) A + cos^(3) A + 3 cos A sin A (sin A + cos A) = m^(3)`
` n+3m sin A cos A = m^(3)`
` [therefore sin^(3) A + cos^(3) A = n and sin A + cos A =m]`
Again , consider sin A + cos A =m
On squaring both sides,
`sin^(2) A + cos^(2) A + 2 sin A cos A = m^(2)`
` sin A cos A = (m^(2) -1)/(2)`
From Eqs. (i) and (ii) , we get
` n + 3m, ((,m^(2) -1))/(2) = m^(3)`
`Rightarrow 2n + 3m^(3) -3m =2m^(2)`
` Rightarrow m^(2) -3m +2n =0`