If `sintheta+costheta=m ,` then prove that `sin^6theta+cos^6theta=(4-3(m^2-1)^2)/4,\ w h e r e\ m^2lt=2.`

Explanation

To show: `sin ^{6} theta+cos ^{6} theta=frac{4-3(m^{2}-1)^{2}}{4}`, where `m^{2} leq 2`

Since, `sin theta+cos theta-m ldots` (i)

`Rightarrow(sin theta+cos theta)^{2}=m^{2}`

`Rightarrow sin ^{2} theta+cos ^{2} theta+2 sin theta cos theta=m^{2}`

`Rightarrow 1+2 sin theta cos theta=m^{2}(because sin ^{2} theta+cos ^{2} theta=1)`

`Rightarrow 2 sin theta cos theta=m^{2}-1`

`Rightarrow sin theta cos theta=frac{m^{2}-1}{2} ldots` (ii)

`therefore` LHS `=sin ^{6} theta+cos ^{2} theta`

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