please answer this question:

Asked by prernamehta | 26th Feb, 2010, 11:38: AM

Expert Answer:

[(sin3A + cos3A)/(sinA + cosA)] + sinAcosA =

[(sinA + cosA)(sin2A+sinAcosA+cos2A)/(sinA + cosA)] + sinAcosA =

...........using a3+b3 = (a+b)(a2-ab+b2)

(sin2A+sinAcosA+cos2A) + sinAcosA =

(1+sinAcosA) + sinAcosA =

1+2sinAcosA =

1+2sinAcosA =

(sin2A+cos2A + 2sinAcosA) = (sinA+cosA)2

Regards,

Team,

TopperLearning.

 

Answered by  | 26th Feb, 2010, 04:32: PM

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