Trigonometry
Asked by khyati11
| 17th Sep, 2008,
03:41: PM
write tanB = sinB / cosB and cross multiply
sinB - n sin2AsinB = n sinA cosA cosB
sinB = nsinA ( sinAsinB+cosAcosB) = nsinA cos(A-B)
now subtract sinAcos(A-B) from both sides
R.H.S = (n-1)sinA cos(A-B)
L.H.S = sinB - sinAcos(A-B)=sinB - sinA (sinAsinB+cosAcosB)
= sinB - sin2A sinB - sinAcosAcosB
=sinB -(1-cos2A) sinB - sinAcosAcosB
=cos2AsinB - sinAcosAcosB
= cosA (cosAsinB - sinA cosB)
= - cosA sin(A-B)
so - cosA sin(A-B) = (n-1)sinA cos(A-B)
tan(A-B) = (1-n) tanA
Answered by
| 18th Sep, 2008,
06:23: PM
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