If 2tanα=3tanβ, prove that tan(α-β)= sin2β 5-cos2β .
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Answer:
Step-by-step explanation:
2tanα=3tanβ
or, tanα=(3/2)tanβ
∴, tan(α-β)=(tanα-tanβ)/(1+tanα.tanβ)
={(3/2)tanβ-tanβ}/{1+(3/2)tanβ.tanβ}
={(3tanβ-2tanβ)/2}/{(2+3tan²β)/2}
=(tanβ/2)×2/(2+3tan²β)
=tanβ/(2+3tan²β)
=(sinβ/cosβ)/{2+3(sin²β/cos²β)}
=(sinβ/cosβ)/{(2cos²β+3sin²β)/cos²β}
=(sinβ/cosβ)×cos²β/{2cos²β+3(1-cos²β)}; [∵,sin²β+cos²β=1]
=sinβcosβ/(2cos²β+3-3cos²β)
=2sinβcosβ/2(3-cos²β)
=sin2β/(6-2cos²β)
=sin2β/(5+1-2cos²β)
=sin2β/{5-(2cos²β-1)}
=sin2β/(5-cos2β) (proved
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