if A+B=C and tanA=k tanB then prove that sin(A-B)=(k-1)/(k+1)sinC
Asked by Aarthi Vishwanathan
| 1st Jun, 2012,
08:55: PM
Expert Answer:
tan A=k tanB ; or tan A\tan B=k
apply componendo & dividendo
(tan A -tanB)\(tanA+tanB)=(k+1)\(k-1)
applying the identities
tan A
tan B =sin (AB)\(cos A cosB)
sin(A-B)=(k-1).sinC\(k+1)
tan A=k tanB ; or tan A\tan B=k
apply componendo & dividendo
(tan A -tanB)\(tanA+tanB)=(k+1)\(k-1)
applying the identities
tan A tan B =sin (AB)\(cos A cosB)
sin(A-B)=(k-1).sinC\(k+1)
Answered by
| 1st Jun, 2012,
11:13: PM
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