hello pavan.i came across these problems while doing a q.paper yeaterday. i find it difficult to solve.pls help me out.pls reply as soon as possible

Asked by  | 18th Apr, 2012, 02:13: PM

Expert Answer:

sin3A+cos3A=(sinA+cosA)(sin2A+cos2A-sinA cosA)
=(sinA+cosA) (1-sinA cosA),                        as sin2A+cos2A=1

Now 1-2cos2A=sin2A+cos2A-2cos2A=sin2A-cos2A=(sinA+cosA)(sinA-cosA)

So, LHS = (sinA+cosA)(1-sinAcosA)/ (sinA+cosA)(sinA-cosA)
               =(1-sinAcosA)/ (sinA-cosA)

Consider RHS now

secA-sinA= (1-sinAcosA)/cosA
and tanA-1= (sinA/cosA) -1=(sinA-cosA)/cosA
So, RHS = (1-sinA cosA)/(sinA-cosA)

Answered by  | 18th Apr, 2012, 03:17: PM

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