(sin A + cos A) ( 1- sinAcosA) = sin³A+ cos³A

Open in App

Solution

L.H.S. = (sinA + cosA)(1 - sinAcosA)
= sinA - sin^2A cosA + cosA - sinA * cos^2A
= sinA - sinA * cos^2A + cosA - sin^2A * cosA
= sinA(1 - cos^2A) + cosA(1 - sin^2A)
= sinA * sin^2A + cosA * cos^2A
= sin^3A + cos^3A
= R.H.S. Proved Ans.

Suggest Corrections

Similar questions

(sin A + cos A) (1 - sin A cos A) = {sin}^{3} A + {cos}^{3} A

\frac{1 - sin A cos A}{cos A (sec A - csc A)} . \frac{{sin}^{2} A - {cos}^{2} A}{{sin}^{3} A + {cos}^{3} A} = sin A

\frac{1 - sinAcosA}{cos A (sec A - cosec A)} . \frac{{sin}^{2} A - {cos}^{2} A}{{sin}^{3} A + {cos}^{3} A} = sin A

\frac{{cos}^{3} A + {sin}^{3} A}{cos A + sin A} + \frac{{cos}^{3} A - {sin}^{3} A}{cos A - sin A}

\frac{S i n^{3} A + c o s^{3} A}{s i n A + c o s A} + \frac{S i n^{3} A - c o s^{3} A}{s i n A - c o s A} = 2

Join BYJU'S Learning Program