Question

if tanA=a/b show that asinA-bcosA/asinA+bcosA=a^2-b^2/a^2+b^2

Answer 1

this is the answer I hope helpful

Answer 2

LHS=(a Sin A -b Cos A)/(a Sin A +b Cos A)
On dividing the whole thing by cos A
We get
LHS
=[(a SinA/cosA)-b(CosA/cosA)]/[(aSinA/cosA)+(bcosA/cosA)]
=[a TanA -b]/[a Tan A +b]. ...sinA/CosA=Tan A
=[(a)(a/b)-b]/[a(a/b)+b]. ....Tan A=(a/b)
=[(a²/b)-b]/[(a²/b)+b].
=[(a²- b²)/b]/[(a²+b²)/b]. ...taking LCM
=(a²-b²)/(a²+b²). ....cancelling out b
=RHS
Hence proved