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यदि `2tan A=3 tan B` तो सिद्ध कीजिए कि `tan(A-B)=(sin 2B)/(5-cos 2B)`

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दिया है -`" "2tan A=3 tan B" …(i)"`
बायाँ पक्ष `=tan (A-B)=(tan A- tanB)/(1+tan A tanB)`
`=(2 tanA-2 tanB)/(2+2 tan A tan B)`
`=(3 tan B - 2 tan B)/(2+3 tan B tan B)=(tan B)/(2+3 tan^(2) B)" "`(समीकरण (i ) का प्रयोग करने पर )
`=(sin B cosB)/(2cos^(2)B+3sin^(2)B)=(2sin B cos B)/(2(2cos^(2)B+3sin^(2)B))`
`=(sin 2B)/(4cos^(2)B+6sin^(2)B)`
`=(sin 2B)/(5cos^(2)B-cos^(2)B+sin^(2)B+5sin^(2)B)`
`=(sin 2B)/(5(cos^(2)B+sin^(2)B)-(cos^(2)B-sin^(2)B))`
`=(sin 2B)/(5-cos 2B)=` बायाँ पक्ष

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