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प्रश्न
Prove that:
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
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उत्तर १
L.H.S. = `(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA)`
= `((cos^3A + sin^3A)(cosA - sinA) + (cos^3A - sin^3A)(cosA + sinA))/(cos^2A - sin^2A)`
= `(cos^4A - cos^3AsinA + sin^3AcosA - sin^4A + cos^4A + cos^3AsinA - sin^3AcosA - sin^4A)/(cos^2A - sin^2A)`
= `(2(cos^4A - sin^4A))/(cos^2A - sin^2A)`
= `(2(cos^2A + sin^2A)2(cos^2A - sin^2A))/(cos^2A - sin^2A)`
= 2(cos2 A + sin2 A)
= 2
= R.H.S. ...(∵ cos2 A + sin2 A = 1)
उत्तर २
L.H.S. = `(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA)`
We apply algebraic identities for the sum and difference of cubes:
a3 − b3 = (a − b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 − ab + b2)
= `((cosA + sinA)(cos^2A - cosA sinA + sin^2A))/((cosA + sinA)) +((cosA - sinA)(cos^2A + cosA sinA + sin^2A))/((cosA - sinA))`
= `(cancel((cosA + sinA))(cos^2A - cosA sinA + sin^2A))/(cancel((cosA + sinA))) +(cancel((cosA - sinA))(cos^2A + cosA sinA + sin^2A))/(cancel((cosA - sinA)))`
= (cos2A − cosA sinA + sin2A) + (cos2A + cosA sinA + sin2A)
= (1 − cosA sinA ) + (1 + cosA sinA) ...(∵ sin2A + cos2A = 1)
= (1 − cosA sinA + 1 + cosA sinA)
= `(1 − cancel(cosA sinA) + 1 + cancel(cosA sinA))`
= 1 + 1
= 2 = R.H.S.
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