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

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)
Hence LHS=RHS