# 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)

Expert's answer

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.

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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