Question

cosA^3+sinA^3/cosA+sinA +cosA^3-sinA^3/cosA-sinA
​

Answer 1

Answer:

We know

${a}^{3} + {b}^{3} = (a + b)( {a}^{2} + {b}^{2} - ab) \\ \\ \frac{a {}^{3} + {b}^{3} }{ {a}^{} + b } = {a}^{2} + {b}^{2} - ab \\ \\ and \\ \\ {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + {b}^{2} + ab) \\ \\ \frac{a {}^{3} - {b}^{3} }{ {a}^{} - b } = {a}^{2} + {b}^{2} + ab$

So your question remains

( think a = cosA and b = sinA )

$( {a}^{2} + {b}^{2} + ab) +( {a}^{2} + {b}^{2} - ab) \\ \\ = 2( {a}^{2} + {b}^{2} ) \\ \\ = 2( \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) ) \\ \\ = 2 \times 1 \\ \\ = 2$

Answer 2

TO PROVE:

Prove that

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

ANSWER:

Taking LHS

Using

a³ + b³ = (a + b)(a² + b² - ab)

Similarly

a³ - b³ = (a - b)(a² + b² + ab)

$\small{ \sf{ \frac{ \cancel{(cosA + sinA)}( {cos}^{2} A + {sin}^{2} A - sinAcosA)}{ \cancel{(cosA + sinA)}} + \frac{ \cancel{(cosA - sinA)}( {cos}^{2}A + {sin}^{2} A + sinAcosA )}{ \cancel{(cosA - sinA)}} }}$

We know that

cos²A + sin²A = 1

$= \small{ \sf{1 + \cancel{ sinAcosA} + 1 - \cancel{sinAcosA}}}$

$= 1 + 1$

$= 2$

LHS = RHS

Hence proved