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Question Number 16807 by sushmitak last updated on 26/Jun/17

(a+2)sin α+(2a−1)cos α=(2a+1) then tan α=?

$$\left({a}+\mathrm{2}\right)\mathrm{sin}\:\alpha+\left(\mathrm{2}{a}−\mathrm{1}\right)\mathrm{cos}\:\alpha=\left(\mathrm{2}{a}+\mathrm{1}\right) \\ $$$$\mathrm{then}\:\mathrm{tan}\:\alpha=? \\ $$

Commented by prakash jain last updated on 26/Jun/17

tan (α/2)=u,sin α=((2u)/(1+u^2 )),cos α=((1−u^2 )/(1+u^2 )) 2(a+2)u+(2a−1)(1−u^2 )=(2a+1)(1+u^2 ) 4au^2 −2(a+2)u+2=0 2au^2 −(a+2)u+1=0 2au^2 −au−2u+1=0 au(2u−1)−(2u−1)=0 (2u−1)(au−1)=0 u=(1/2),u=(1/a) tan α=((2u)/(1−u^2 ))=(4/3) tan α=((2a)/(a^2 −1))

$$\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}={u},\mathrm{sin}\:\alpha=\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} },\mathrm{cos}\:\alpha=\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\mathrm{2}\left({a}+\mathrm{2}\right){u}+\left(\mathrm{2}{a}−\mathrm{1}\right)\left(\mathrm{1}−{u}^{\mathrm{2}} \right)=\left(\mathrm{2}{a}+\mathrm{1}\right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right) \\ $$$$\mathrm{4}{au}^{\mathrm{2}} −\mathrm{2}\left({a}+\mathrm{2}\right){u}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}{au}^{\mathrm{2}} −\left({a}+\mathrm{2}\right){u}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}{au}^{\mathrm{2}} −{au}−\mathrm{2}{u}+\mathrm{1}=\mathrm{0} \\ $$$${au}\left(\mathrm{2}{u}−\mathrm{1}\right)−\left(\mathrm{2}{u}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}{u}−\mathrm{1}\right)\left({au}−\mathrm{1}\right)=\mathrm{0} \\ $$$${u}=\frac{\mathrm{1}}{\mathrm{2}},{u}=\frac{\mathrm{1}}{{a}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}{u}}{\mathrm{1}−{u}^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}{a}}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$

Answered by ajfour last updated on 26/Jun/17

(a+2)sin α= (1+cos α)+2a(1−cos α) 2(a+2)sin (α/2)cos (α/2) = 2cos^2 (α/2)+4asin^2 (α/2) dividing by 2cos^2 (α/2): (a+2)tan (α/2)=1+2a tan^2 (α/2) 2atan^2 (α/2)−(a+2)tan (α/2)+1=0 ⇒ atan (α/2)[2tan (α/2)−1] −[2tan (α/2)−1]=0 [2tan (α/2)−1][atan (α/2)−1]=0 ⇒ tan (α/2)=(1/2), (1/a) As tan α=((2tan α)/(1−tan^2 α)) tan α = ((2(1/2))/(1−(1/4))) = (4/3) or tan α = ((2(1/a))/(1−(1/a)^2 )) = ((2a)/(a^2 −1)) .

$$\:\left(\mathrm{a}+\mathrm{2}\right)\mathrm{sin}\:\alpha=\:\left(\mathrm{1}+\mathrm{cos}\:\alpha\right)+\mathrm{2a}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right) \\ $$$$\mathrm{2}\left(\mathrm{a}+\mathrm{2}\right)\mathrm{sin}\:\left(\alpha/\mathrm{2}\right)\mathrm{cos}\:\left(\alpha/\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{2cos}\:^{\mathrm{2}} \left(\alpha/\mathrm{2}\right)+\mathrm{4asin}\:^{\mathrm{2}} \left(\alpha/\mathrm{2}\right) \\ $$$$\:\mathrm{dividing}\:\mathrm{by}\:\mathrm{2cos}\:^{\mathrm{2}} \left(\alpha/\mathrm{2}\right): \\ $$$$\:\left(\mathrm{a}+\mathrm{2}\right)\mathrm{tan}\:\left(\alpha/\mathrm{2}\right)=\mathrm{1}+\mathrm{2a}\:\mathrm{tan}\:^{\mathrm{2}} \left(\alpha/\mathrm{2}\right) \\ $$$$\:\mathrm{2atan}\:^{\mathrm{2}} \left(\alpha/\mathrm{2}\right)−\left(\mathrm{a}+\mathrm{2}\right)\mathrm{tan}\:\left(\alpha/\mathrm{2}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{atan}\:\left(\alpha/\mathrm{2}\right)\left[\mathrm{2tan}\:\left(\alpha/\mathrm{2}\right)−\mathrm{1}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left[\mathrm{2tan}\:\left(\alpha/\mathrm{2}\right)−\mathrm{1}\right]=\mathrm{0} \\ $$$$\:\left[\mathrm{2tan}\:\left(\alpha/\mathrm{2}\right)−\mathrm{1}\right]\left[\mathrm{atan}\:\left(\alpha/\mathrm{2}\right)−\mathrm{1}\right]=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\left(\alpha/\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{2}},\:\:\frac{\mathrm{1}}{\mathrm{a}} \\ $$$$\:\mathrm{As}\:\:\:\mathrm{tan}\:\alpha=\frac{\mathrm{2tan}\:\alpha}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \alpha} \\ $$$$\:\:\mathrm{tan}\:\alpha\:=\:\frac{\mathrm{2}\left(\mathrm{1}/\mathrm{2}\right)}{\mathrm{1}−\left(\mathrm{1}/\mathrm{4}\right)}\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{or} \\ $$$$\:\mathrm{tan}\:\alpha\:=\:\frac{\mathrm{2}\left(\mathrm{1}/\mathrm{a}\right)}{\mathrm{1}−\left(\mathrm{1}/\mathrm{a}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{2a}}{\mathrm{a}^{\mathrm{2}} −\mathrm{1}}\:. \\ $$$$ \\ $$

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