Question

2 tanA=3 tan B Prove that tan(A-B)=sin 2B/5-cos2B

Answer 1

\[\tan( A - B) = \dfrac{\tan A - \tan
B}{1 + \tan A \tan B}\]Since \(\tan A = \frac{3}{2}\tan B\), you can plug it in.\[= \dfrac{\frac{1}{2}\tan B}{1 + \frac{3}{2}\tan ^2 B} = \dfrac{\tan B}{2 + 3 \tan^2 B}\]

Answer 2

tanA={3 tan B/ 2 } ------------------(1)

Prove that: tan(A-B)=sin 2B/5-cos2B

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 - { cos22B}]

= sin2B / 5 - cos22B

LHS = RHS (Hence proved)