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In $Δ A B C \tan A + \tan B + \tan C = 6 a n d \tan A \cdot \tan B = 2 then the possible value of \sin^{2} A : \sin^{2} B : \sin^{2} C =$

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8:9:5

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5:9:8

5:8:7

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Correct option is B

tanA+tanB+tanC=tanAtanBtanC=6⇒2tanC=6 ∵tanAtanB=2⇒tanC=3∴tanA+tanB=6-3⇒tanA+tanB=3 ⇒tanA=2 and tanB=1⇒sin2A:sin2B:sin2C=tan2A1+tan2A:tan2B1+tan2B:tan2C1+tan2C =45:12:910 =8:5:9

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In ΔABC tanA+tanB+tanC=6 and tanA·tanB=2 then the possible value of sin2A:sin2B:sin2C=

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

In $Δ A B C \tan A + \tan B + \tan C = 6 a n d \tan A \cdot \tan B = 2 then the possible value of \sin^{2} A : \sin^{2} B : \sin^{2} C =$