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Question Number 21453 by Joel577 last updated on 24/Sep/17

If (1 + n)sin 2θ + (1 − n)cos 2θ = 1 + n find tan 2θ

$$\mathrm{If}\: \\ $$$$\left(\mathrm{1}\:+\:{n}\right)\mathrm{sin}\:\mathrm{2}\theta\:+\:\left(\mathrm{1}\:−\:{n}\right)\mathrm{cos}\:\mathrm{2}\theta\:=\:\mathrm{1}\:+\:{n} \\ $$$$\mathrm{find}\:\mathrm{tan}\:\mathrm{2}\theta \\ $$

Answered by $@ty@m last updated on 24/Sep/17

(1−n)cos 2θ=(1+n)(1−sin 2θ) ((1−n)/(1+n))=((1−sin 2θ)/(cos 2θ))=(((cos θ−sin θ)^2 )/(cos^2 θ−sin^2 θ)) ((1−n)/(1+n))=((cos θ−sin θ)/(cos θ+sin θ)) ((−2n)/2)=((−2sinθ )/(2cos θ)), by componendo & dividendo tanθ=n ∴ tan2θ=((2tan θ)/(1−tan^2 θ)) ⇒ tan2θ=((2n)/(1−n^2 )) Ans.

$$\left(\mathrm{1}−{n}\right)\mathrm{cos}\:\mathrm{2}\theta=\left(\mathrm{1}+{n}\right)\left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$$\frac{\mathrm{1}−{n}}{\mathrm{1}+{n}}=\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{cos}\:\mathrm{2}\theta}=\frac{\left(\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$$\frac{\mathrm{1}−{n}}{\mathrm{1}+{n}}=\frac{\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta} \\ $$$$\frac{−\mathrm{2}{n}}{\mathrm{2}}=\frac{−\mathrm{2sin}\theta\:}{\mathrm{2cos}\:\theta},\:{by}\:{componendo}\:\&\:{dividendo} \\ $$$$\mathrm{tan}\theta={n} \\ $$$$\therefore\:\mathrm{tan2}\theta=\frac{\mathrm{2tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta}\: \\ $$$$\Rightarrow\:\mathrm{tan2}\theta=\frac{\mathrm{2}{n}}{\mathrm{1}−{n}^{\mathrm{2}} }\:{Ans}. \\ $$$$ \\ $$

Answered by sma3l2996 last updated on 24/Sep/17

(1+n)sin2θ+(1−n)cos2θ=1+n ⇔(1+n)tan2θ+1−n=((1+n)/(cos2θ)) we have (1/(cosa))=(√(1+tan^2 a)) so (1+n)tan2θ+1−n=(1+n)(√(1+tan^2 2θ)) (1+n)^2 tan^2 2θ+(1−n)^2 +2(1+n)(1−n)tan2θ=(1+n)^2 (1+tan^2 2θ) (1−n)^2 +2(1−n^2 )tan2θ=(1+n)^2 2(1−n^2 )tan2θ=(1+n)^2 −(1−n)^2 =4n tan2θ=((2n)/(1−n^2 ))

$$\left(\mathrm{1}+{n}\right){sin}\mathrm{2}\theta+\left(\mathrm{1}−{n}\right){cos}\mathrm{2}\theta=\mathrm{1}+{n} \\ $$$$\Leftrightarrow\left(\mathrm{1}+{n}\right){tan}\mathrm{2}\theta+\mathrm{1}−{n}=\frac{\mathrm{1}+{n}}{{cos}\mathrm{2}\theta} \\ $$$${we}\:{have}\:\frac{\mathrm{1}}{{cosa}}=\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {a}} \\ $$$${so}\:\:\left(\mathrm{1}+{n}\right){tan}\mathrm{2}\theta+\mathrm{1}−{n}=\left(\mathrm{1}+{n}\right)\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$\left(\mathrm{1}+{n}\right)^{\mathrm{2}} {tan}^{\mathrm{2}} \mathrm{2}\theta+\left(\mathrm{1}−{n}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}+{n}\right)\left(\mathrm{1}−{n}\right){tan}\mathrm{2}\theta=\left(\mathrm{1}+{n}\right)^{\mathrm{2}} \left(\mathrm{1}+{tan}^{\mathrm{2}} \mathrm{2}\theta\right) \\ $$$$\left(\mathrm{1}−{n}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}−{n}^{\mathrm{2}} \right){tan}\mathrm{2}\theta=\left(\mathrm{1}+{n}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{1}−{n}^{\mathrm{2}} \right){tan}\mathrm{2}\theta=\left(\mathrm{1}+{n}\right)^{\mathrm{2}} −\left(\mathrm{1}−{n}\right)^{\mathrm{2}} =\mathrm{4}{n} \\ $$$${tan}\mathrm{2}\theta=\frac{\mathrm{2}{n}}{\mathrm{1}−{n}^{\mathrm{2}} } \\ $$

Answered by $@ty@m last updated on 24/Sep/17

Another solution: (1+n)((2tanθ )/(1+tan^2 θ ))+(1−n)((1−tan^2 θ )/(1+tan^2 θ ))=1+n 2(1+n)y+(1−n)(1−y^2 )=(1+n)(1+y^2 ), where y=tanθ 2(1+n)y+1−n−y^2 +ny^2 =1+n+y^2 +ny^2 2(1+n)y=2(n+y^2 ) y^2 −(1+n)y+n=0 y=((1+n±(√((1+n)^2 −4n)))/2) y=((1+n±(1−n))/2) y=1, n tan 2θ=∞, ((2n)/(1−n^2 ))

$${Another}\:{solution}: \\ $$$$\left(\mathrm{1}+{n}\right)\frac{\mathrm{2tan}\theta\:}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\:}+\left(\mathrm{1}−{n}\right)\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \theta\:}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\:}=\mathrm{1}+{n} \\ $$$$\mathrm{2}\left(\mathrm{1}+{n}\right){y}+\left(\mathrm{1}−{n}\right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right)=\left(\mathrm{1}+{n}\right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right),\:{where}\:{y}=\mathrm{tan}\theta\: \\ $$$$\mathrm{2}\left(\mathrm{1}+{n}\right){y}+\mathrm{1}−{n}−{y}^{\mathrm{2}} +{ny}^{\mathrm{2}} =\mathrm{1}+{n}+{y}^{\mathrm{2}} +{ny}^{\mathrm{2}} \\ $$$$\mathrm{2}\left(\mathrm{1}+{n}\right){y}=\mathrm{2}\left({n}+{y}^{\mathrm{2}} \right) \\ $$$${y}^{\mathrm{2}} −\left(\mathrm{1}+{n}\right){y}+{n}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{1}+{n}\pm\sqrt{\left(\mathrm{1}+{n}\right)^{\mathrm{2}} −\mathrm{4}{n}}}{\mathrm{2}} \\ $$$${y}=\frac{\mathrm{1}+{n}\pm\left(\mathrm{1}−{n}\right)}{\mathrm{2}} \\ $$$${y}=\mathrm{1},\:{n} \\ $$$$\mathrm{tan}\:\mathrm{2}\theta=\infty,\:\frac{\mathrm{2}{n}}{\mathrm{1}−{n}^{\mathrm{2}} } \\ $$

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