The value of frac{sin ^{3} A+cos ^{3} A}{sin | vedclass

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(C) We use the algebraic identities $a^3 + b^3 = (a + b)(a^2 + b^2 - ab)$ and $a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$.Given expression: $\frac{\sin ^{3}

Vedclass · Accepted Answer

$2$

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(C) We use the algebraic identities $a^3 + b^3 = (a + b)(a^2 + b^2 - ab)$ and $a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$.Given expression: $\frac{\sin ^{3} A+\cos ^{3} A}{\sin A+\cos A}+\frac{\cos ^{3} A-\sin ^{3} A}{\cos A-\sin A}$$= \frac{(\sin A + \cos A)(\sin^2 A + \cos^2 A - \sin A \cos A)}{\sin A + \cos A} + \frac{(\cos A - \sin A)(\cos^2 A + \sin^2 A + \sin A \cos A)}{\cos A - \sin A}$$= (\sin^2 A + \cos^2 A - \sin A \cos A) + (\cos^2 A + \sin^2 A + \sin A \cos A)$Since $\sin^2 A + \cos^2 A = 1$,we have:$= (1 - \sin A \cos A) + (1 + \sin A \cos A)$$= 1 - \sin A \cos A + 1 + \sin A \cos A$$= 2$

The value of $\frac{\sin ^{3} A+\cos ^{3} A}{\sin A+\cos A}+\frac{\cos ^{3} A-\sin ^{3} A}{\cos A-\sin A}$ is

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