Question

cos3A+sin3A/cosA+cos3A-sin3A/cosA-sinA=2

Answer 1

Answer:

$\red{\frac{cos^{3}A+sin^{3}A}{cosA+sinA}+\frac{cos^{3}A-sin^{3}A}{cosA-sinA}}\green{=2}$

Step-by-step explanation:

$LHS = \frac{cos^{3}A+sin^{3}A}{cosA+sinA}+\frac{cos^{3}A-sin^{3}A}{cosA-sinA}$

$=\frac{(cosA+sinA)(cos^{2}A-cosAsinA+sin^{2}A}{(cosA+sinA)}+\frac{(cosA-sinA)(cos^{2}A+cosAsinA+sin^{2}A}{(cosA-sinA)}$

$=cos^{2}A-cosAsinA+sin^{2}A+cos^{2}A+cosAsinA+sin^{2}A$

$= 2(cos^{2}A+sin^{2}A)$

$= 2\times 1$

$\boxed {\pink {cos^{2}A+sin^{2}A=1}}$

$= RHS$

Therefore.,

$\red{\frac{cos^{3}A+sin^{3}A}{cosA+sinA}+\frac{cos^{3}A-sin^{3}A}{cosA-sinA}}\green{=2}$

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