1.

1+cosA+sinA/1+cosA-sinA=1+sinA/cosA

Answer»

(sinA+1-cosA)/(cosA-1+sinA) =(1+sinA)/cosAL.H.S.=[sinA+(1-cosA)]/[sinA-(1-cosA)]MULTIPLYING by [sinA+(1-cosA)] in NR and DR.=[sinA+(1-cosA)]^2/[sin^2A-(1-cosA)^2]=[sin^2A+2.sinA.(1-cosA)+(1-cosA)^2]/[(1-cos^2A)-(1-cosA)^2]=[(1-cos^2A)+2.sinA.(1-cosA) +(1-cosA)^2]/[(1-cosA)(1+cosA) - (1-cosA)^2].=(1-cosA)[1+cosA+2sinA+1-cosA]/(1-cosA)[1+cosA-1+cosA].=[2+2sinA]/[2.cosA].=2(1+sinA)/2.cosA.= (1+sinA)/cosA , proved.



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