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Question Number 62983 by hovea cw last updated on 27/Jun/19

If tan 2θ tan θ = 1, then θ =

$$\mathrm{If}\:\:\:\mathrm{tan}\:\mathrm{2}\theta\:\mathrm{tan}\:\theta\:=\:\mathrm{1},\:\mathrm{then}\:\theta\:= \\ $$

Answered by Hope last updated on 27/Jun/19

tan2θ=cotθ=tan((π/2)−θ) formula [tanα=tanβ α=nπ+β] so 2θ=nπ+((π/2)−θ) 3θ=(((2n+1)π)/2)→ θ=(((2n+1)π)/6)

$${tan}\mathrm{2}\theta={cot}\theta={tan}\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$${formula} \\ $$$$\left[{tan}\alpha={tan}\beta\:\:\:\alpha={n}\pi+\beta\right] \\ $$$${so}\:\mathrm{2}\theta={n}\pi+\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$$\mathrm{3}\theta=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{2}}\rightarrow\:\theta=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{6}} \\ $$

Commented by mr W last updated on 27/Jun/19

please check sir! n=1: θ=(π/2) ⇒ but tan (π/2) → ∞ !

$${please}\:{check}\:{sir}! \\ $$$${n}=\mathrm{1}:\:\theta=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{but}\:\mathrm{tan}\:\frac{\pi}{\mathrm{2}}\:\rightarrow\:\infty\:! \\ $$

Commented by Hope last updated on 27/Jun/19

yes sir...you r right...

$${yes}\:{sir}...{you}\:{r}\:{right}... \\ $$

Answered by Hope last updated on 27/Jun/19

anothdr way a=tanθ ((2a)/(1−a^2 ))×a=1 2a^2 =1−a^2 a^2 =(1/3)→a=±(1/(√3)) tanθ=tan(π/6) θ=nπ+(π/6) tanθ=−(1/(√3))=tan(π−(π/6)) θ=nπ+((5π)/6)

$${anothdr}\:{way} \\ $$$${a}={tan}\theta \\ $$$$\frac{\mathrm{2}{a}}{\mathrm{1}−{a}^{\mathrm{2}} }×{a}=\mathrm{1} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} =\mathrm{1}−{a}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\rightarrow{a}=\pm\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$${tan}\theta={tan}\frac{\pi}{\mathrm{6}} \\ $$$$\theta={n}\pi+\frac{\pi}{\mathrm{6}} \\ $$$${tan}\theta=−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}={tan}\left(\pi−\frac{\pi}{\mathrm{6}}\right) \\ $$$$\theta={n}\pi+\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$

Commented by mr W last updated on 27/Jun/19

good sir! that is θ=nπ±(π/6)

$${good}\:{sir}! \\ $$$${that}\:{is}\:\theta={n}\pi\pm\frac{\pi}{\mathrm{6}} \\ $$

Answered by MJS last updated on 27/Jun/19

tan 2θ =((2sin θ cos θ)/(2cos^2 θ −1)) ((2sin θ cos θ)/(2cos^2 θ −1))×((sin θ)/(cos θ))=1 ((2sin^2 θ)/(2cos^2 θ −1))=1 method 1 2sin^2 θ =2cos^2 θ −1 sin^2 θ −cos^2 θ =−(1/2) 1−2cos^2 θ =−(1/2) cos^2 θ =(3/4) cos θ =±((√3)/2) method 2 ((1−cos 2θ)/(cos 2θ))=1 cos 2θ =(1/2) ...

$$\mathrm{tan}\:\mathrm{2}\theta\:=\frac{\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{2cos}^{\mathrm{2}} \:\theta\:−\mathrm{1}} \\ $$$$\frac{\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{2cos}^{\mathrm{2}} \:\theta\:−\mathrm{1}}×\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}=\mathrm{1} \\ $$$$\frac{\mathrm{2sin}^{\mathrm{2}} \:\theta}{\mathrm{2cos}^{\mathrm{2}} \:\theta\:−\mathrm{1}}=\mathrm{1} \\ $$$$\mathrm{method}\:\mathrm{1} \\ $$$$\mathrm{2sin}^{\mathrm{2}} \:\theta\:=\mathrm{2cos}^{\mathrm{2}} \:\theta\:−\mathrm{1} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\theta\:−\mathrm{cos}^{\mathrm{2}} \:\theta\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}−\mathrm{2cos}^{\mathrm{2}} \:\theta\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\theta\:=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\theta\:=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{method}\:\mathrm{2} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\mathrm{2}\theta}=\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$... \\ $$

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