Trigonometry

Asked by khyati11 | 17th Sep, 2008, 03:41: PM

Expert Answer:

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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