Answer to Question #11656 in Trigonometry for Avruti
2012-07-10T07:53:21-04:00
if 2tan(alpha)=3tan(beta), prove that tan(alpha-beta)=sin(2beta)/5-cos(2beta)
1
2012-07-10T10:38:28-0400
tanA={3 tan B/ 2 } (1) LHS : tan (A-B) = tanA – tanB / 1 +tanA tanB = (3 tanB/2) – tanB / 1 + tanB *3tanB/2 [From(1)] = [(3tanB – 2tanB)/2 ] / [(2 +3 tan²B)/2 ] = { [(tanB)/2 ]* 2 } / [(2 +3 tan²B) ] = (tanB) / (2 + tan²B) = (sinB/cosB) / (2 +{ 3sin²B/cos²B} ) = (sinB/cosB) / { ([2 cos²B +3sin²B]/cos²B ) } = (sinB* cos²B /cosB) / [2 cos²B +3sin²B] = (sinB*cosB) / [2 cos2B +3sin2B] Multiplying the NUMERATOR and DENOMINATOR by “2”: = { 2(sinB*cosB) }/ { 2[2 cos²B +3sin²B]} = {sin2B} /[ 4cos²B +6sin²B] (2sinB*cosB = sin2B) = {sin2B} /[ 4cos²B +6(1-cos²B) ] = {sin2B} /[ 6-2cos²B) ] = {sin2B} /[ (1+5-2cos²B) ] = {sin2B} /[ 5+{1-2cos²B}] = {sin2B} /[ 5 - { 2cos²B-1}] (2cos²B-1 = cos2x) = {sin2B} /[ 5 - { cos2B}] = sin2B / 5 - cos2B LHS = RHS, hence proved.
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