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Question Number 48849 by vajpaithegrate@gmail.com last updated on 29/Nov/18

in (0 π)the number of solutions of the equation tanθ+tan2θ+tan3θ=tanθtan2θtan3θ is ans:2

$$\mathrm{in}\:\left(\mathrm{0}\:\pi\right)\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{tan}\theta+\mathrm{tan2}\theta+\mathrm{tan3}\theta=\mathrm{tan}\theta\mathrm{tan2}\theta\mathrm{tan3}\theta \\ $$$$\mathrm{is} \\ $$$$\mathrm{ans}:\mathrm{2} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Nov/18

tan3θ=((tan2θ+tanθ)/(1−tan2θtanθ)) tan3θ−tanθtan2θtan3θ=tan2θ+tanθ tan3θ−tan2θ−tanθ=tanθtan2θtan3θ so tanθ+tan2θ+tan3θ=tan3θ−tan2θ−tanθ 2(tanθ+tan2θ)=0 t+((2t)/(1−t^2 ))=0 t=tanθ t−t^3 +2t=0 3t−t^3 =0 t(3−t^2 )= t=0 t^2 =3 t=±(√3) given interval is (0,π) so solution are t=±(√3) that is tanθ=±(√3) so θ=60^o and 120^o θ=(π/3),((2π)/3) pls check...

$${tan}\mathrm{3}\theta=\frac{{tan}\mathrm{2}\theta+{tan}\theta}{\mathrm{1}−{tan}\mathrm{2}\theta{tan}\theta} \\ $$$${tan}\mathrm{3}\theta−{tan}\theta{tan}\mathrm{2}\theta{tan}\mathrm{3}\theta={tan}\mathrm{2}\theta+{tan}\theta \\ $$$${tan}\mathrm{3}\theta−{tan}\mathrm{2}\theta−{tan}\theta={tan}\theta{tan}\mathrm{2}\theta{tan}\mathrm{3}\theta \\ $$$${so} \\ $$$${tan}\theta+{tan}\mathrm{2}\theta+{tan}\mathrm{3}\theta={tan}\mathrm{3}\theta−{tan}\mathrm{2}\theta−{tan}\theta \\ $$$$\mathrm{2}\left({tan}\theta+{tan}\mathrm{2}\theta\right)=\mathrm{0} \\ $$$${t}+\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }=\mathrm{0}\:\:\:\:{t}={tan}\theta \\ $$$${t}−{t}^{\mathrm{3}} +\mathrm{2}{t}=\mathrm{0} \\ $$$$\mathrm{3}{t}−{t}^{\mathrm{3}} =\mathrm{0} \\ $$$${t}\left(\mathrm{3}−{t}^{\mathrm{2}} \right)= \\ $$$${t}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} =\mathrm{3}\:\:\:{t}=\pm\sqrt{\mathrm{3}}\: \\ $$$$ \\ $$$${given}\:{interval}\:{is}\:\left(\mathrm{0},\pi\right) \\ $$$${so}\:{solution}\:{are} \\ $$$${t}=\pm\sqrt{\mathrm{3}}\: \\ $$$${that}\:{is}\:{tan}\theta=\pm\sqrt{\mathrm{3}}\: \\ $$$${so}\:\theta=\mathrm{60}^{{o}} \:{and}\:\mathrm{120}^{{o}} \\ $$$$\theta=\frac{\pi}{\mathrm{3}},\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$${pls}\:{check}... \\ $$$$ \\ $$

Commented by vajpaithegrate@gmail.com last updated on 29/Nov/18

tnq sir

$$\mathrm{tnq}\:\mathrm{sir} \\ $$

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