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Question

If tanθ+tan4θ+tan7θ=tanθtan4θtan7θ, then θ=

A
θ=nπ4
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B
θ=nπ7
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C
θ=nπ12;n6(2k+1)(n,kI)
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D
θ=nπ
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Solution

The correct option is C θ=nπ12;n6(2k+1)(n,kI)
We know tan(A+B+C)=tanA+tanB+tanCtanAtanBtanC1tanAtanBtanBtanCtanAtanC
let A=θ,B=4θ,C=7θ
then numerator tanθ+tan4θ+tan7θtanθtan4θtan7θ
So to solve if numberator =0
then tan(θ+4θ+7θ)=tan12θ=0
or 12θ=nπ
θ=11π12

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