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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$2$

Vedclass · Answer

(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

Similar Questions

If $\sin A + \sin B = C$ and $\cos A + \cos B = D$,then the value of $\sin (A + B) = $

In a $\Delta ABC$,$\angle B = \frac{\pi}{3}$,$\angle C = \frac{\pi}{4}$ and $D$ divides $BC$ internally in the ratio $1:3$,then $\frac{\sin \angle BAD}{\sin \angle CAD}$ is equal to

$\tan 20^\circ + \tan 40^\circ + \sqrt{3} \tan 20^\circ \tan 40^\circ = $

If $\cos x + \cos y + \cos \alpha = 0$ and $\sin x + \sin y + \sin \alpha = 0,$ then $\cot \left( \frac{x + y}{2} \right) = $

If for real values of $x$,$\cos \theta = x + \frac{1}{x}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module