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# Answer to Question #11656 in Trigonometry for Avruti

Question #11656
if 2tan(alpha)=3tan(beta), prove that tan(alpha-beta)=sin(2beta)/5-cos(2beta)
tanA={3 tan B/ 2 } (1)

LHS : tan (A-B)
= tanA &ndash; tanB / 1 +tanA tanB
= (3 tanB/2) &ndash; tanB / 1 + tanB *3tanB/2 [From(1)]
= [(3tanB &ndash; 2tanB)/2 ] / [(2 +3 tan&sup2;B)/2 ]
= { [(tanB)/2 ]* 2 } / [(2 +3 tan&sup2;B) ]
= (tanB) / (2 + tan&sup2;B)
= (sinB/cosB) / (2 +{ 3sin&sup2;B/cos&sup2;B} )
= (sinB/cosB) / { ([2 cos&sup2;B +3sin&sup2;B]/cos&sup2;B ) }
= (sinB* cos&sup2;B /cosB) / [2 cos&sup2;B +3sin&sup2;B]
= (sinB*cosB) / [2 cos2B +3sin2B]

Multiplying the NUMERATOR and DENOMINATOR by &ldquo;2&rdquo;:

= { 2(sinB*cosB) }/ { 2[2 cos&sup2;B +3sin&sup2;B]}
= {sin2B} /[ 4cos&sup2;B +6sin&sup2;B] (2sinB*cosB = sin2B)
= {sin2B} /[ 4cos&sup2;B +6(1-cos&sup2;B) ]
= {sin2B} /[ 6-2cos&sup2;B) ]
= {sin2B} /[ (1+5-2cos&sup2;B) ]
= {sin2B} /[ 5+{1-2cos&sup2;B}]
= {sin2B} /[ 5 - { 2cos&sup2;B-1}] (2cos&sup2;B-1 = cos2x)
= {sin2B} /[ 5 - { cos2B}]
= sin2B / 5 - cos2B

LHS = RHS, hence proved.

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