यदि `sin theta+cos theta=x,` तो सिद्ध कीजिए कि `sin^(6)theta+cos^(6)theta=1-(3)/(4)(x^(2)-1)^2,x^(2) le 2`

Explanation

दायाँ पक्ष `=1-(3)/(4)(x^(2)-1)^(2)=1-(3)/(4)[(sin theta=cos theta)^(2)-1]^(2)=1-(3)/(4)[sin^(2)theta+cos^(2)theta+2sin theta cos theta-1]^(2)`
`=1-3sin^(2)theta cos^(2) theta=(sin^(2)theta+cos^(2)theta)^(3)-3sin^(2)thetacos^(2)theta`
`=(sin^(2)theta+cos^(2)theta)^(3)-3sin^(2)thetacos^(2)theta(sin^(2)theta+cos^(2)theta)`
`=(sin^(2)theta)^(3)+(cos^(2)theta)^(3)=sin^(6)theta+cos^(6)theta`