Question

10. If TanA=
1-cos B
sin B
then tan2A=​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

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Step-by-step explanation:

∵cosAcosBcosC=

∵cosAcosBcosC= 8

∵cosAcosBcosC= 83

∵cosAcosBcosC= 83

∵cosAcosBcosC= 83 −1

∵cosAcosBcosC= 83 −1

∵cosAcosBcosC= 83 −1 and sinAsinBsinC=

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 8

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC=

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −1

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3 As in △ABC,A+B+C=π

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3 As in △ABC,A+B+C=π∴tanAtanBtanC=tanA+tanB+tanC=

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3 As in △ABC,A+B+C=π∴tanAtanBtanC=tanA+tanB+tanC= 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3 As in △ABC,A+B+C=π∴tanAtanBtanC=tanA+tanB+tanC= 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3 As in △ABC,A+B+C=π∴tanAtanBtanC=tanA+tanB+tanC= 3 −1

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3 As in △ABC,A+B+C=π∴tanAtanBtanC=tanA+tanB+tanC= 3 −13+

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3 As in △ABC,A+B+C=π∴tanAtanBtanC=tanA+tanB+tanC= 3 −13+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3 As in △ABC,A+B+C=π∴tanAtanBtanC=tanA+tanB+tanC= 3 −13+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3 As in △ABC,A+B+C=π∴tanAtanBtanC=tanA+tanB+tanC= 3 −13+ 3

∵cosAcosBcosC= 83 −1 and sinAsinBsinC= 83+ 3 ∴tanAtanBtanC= 3 −13+ 3 As in △ABC,A+B+C=π∴tanAtanBtanC=tanA+tanB+tanC= 3 −13+ 3