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Mathematics

Prove that :

cos3A+sin3Acos A+ sin A+cos3A−sin3Acos A - sin A\dfrac{\text{cos}^3 A + \text{sin}^3 A}{\text{cos A+ sin A}} + \dfrac{\text{cos}^3 A - \text{sin}^3 A}{\text{cos A - sin A}} = 2

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Answer

Solving L.H.S. of the equation :

⇒(cos3A+sin3A)(cos A - sin A)+(cos3A−sin3A)(cos A + sin A)(cos A + sin A)(cos A - sin A)⇒cos4A−cos3A sin A+cos A sin3A− sin4A+cos4A+cos3A sin A− sin3A cos A−sin4Acos2A−sin2A⇒2 cos4A−2 sin4Acos2A−sin2A⇒2(cos4A− sin4A)cos2A−sin2A⇒2(cos2A−sin2A)(cos2A+sin2A)cos2A−sin2A⇒2 (cos2A+sin2A).\Rightarrow \dfrac{(\text{cos}^3 A + \text{sin}^3 A)(\text{cos A - sin A}) + (\text{cos}^3 A - \text{sin}^3 A)(\text{cos A + sin A})}{\text{(cos A + sin A)(cos A - sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos}^4 A - \text{cos}^3 A\text{ sin A} + \text{cos A sin}^3 A - \text{ sin}^4 A + \text{cos}^4 A + \text{cos}^3 A\text{ sin A} - \text{ sin}^3 A\text{ cos A} - \text{sin}^4 A}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 cos}^4 A - \text{2 sin}^4 A}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2(cos}^4 A - \text{ sin}^4 A)}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2(cos}^2 A - \text{sin}^2 A)\text{(cos}^2 A + \text{sin}^2 A)}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \text{2 (cos}^2 A + \text{sin}^2 A).

By formula,

cos2 A + sin2 A = 1

⇒ 2 × 1 = 2.

Since, L.H.S. = R.H.S.

Hence, proved that cos3A+sin3Acos A+ sin A+cos3A−sin3Acos A - sin A\dfrac{\text{cos}^3 A + \text{sin}^3 A}{\text{cos A+ sin A}} + \dfrac{\text{cos}^3 A - \text{sin}^3 A}{\text{cos A - sin A}} = 2.

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