Math, asked by jil, 1 year ago

(1+ cotA + tanA )(sinA - cosA) = sinAtanA - cotAcosA


annumalekar: wat s wrng?could u plz tell me
annumalekar: my ans was correct

Answers

Answered by Anonymous
7
from RHS
sinAtanA - cotAcosA

 {sin^{2} A}/{cosA} - cos^{2}A/sinA = (sin^{3}A - cos^{3}A)/sinAcosA=(sinA - cosA)(sin^{2}A + cos^{2}A + sinAcosA)/sinAcosA = (sinA - cosA)[(sin^{2}A + cos^{2}A + sinAcosA)/sinAcosA ] =  (sinA - cosA)[(sin^{2}A)/sinAcosA + (cos^{2}A)/sinAcosA + 1] =  (sinA - cosA)(1+ cotA + tanA)[/tex[tex]PROVED.

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