Answers

2015-11-06T18:09:51+05:30
A^3+b^3=(a+b)(a^2-ab+b^2) a^3-b^3=(a-b)(a^2+ab+b^2) so cos^3A+sin^3A=(cosA+sinA)(cos^2A-cosAsinA+sin^2A) so [cos^3A+sin^3A]/[cosA+sinA]=1-sinAcosA......................1 sin^2A+cos^2A=1 similarly cos^A-sin^2A=(cosA-sinA)(cos^2A+sinAcosA+sin^2A) so [cos^3A-sin^3A]/[cosA-sinA]=1+sinAcosA.............................2 so adding both equation 1-sinAcosA+1+sinAcosA 2 LHS=RHS
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