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Question Number 222279 by fantastic last updated on 21/Jun/25
(a^2 −b^2 )sin θ+2abcos θ=a^2 +b^2 tan θ=??
$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{sin}\:\theta+\mathrm{2}{ab}\mathrm{cos}\:\theta={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=?? \\ $$
Answered by mr W last updated on 21/Jun/25
((a^2 −b^2 )/(a^2 +b^2 )) sin θ+((2ab)/(a^2 +b^2 )) cos θ=1 since (((a^2 −b^2 )/(a^2 +b^2 )))^2 +(((2ab)/(a^2 +b^2 )))^2 =1, ((a^2 −b^2 )/(a^2 +b^2 ))=sin α, ((2ab)/(a^2 +b^2 ))=cos α sin α sin θ+cos α cos θ=1 cos (θ−α)=1 ⇒θ−α=2kπ ⇒θ=2kπ+α ⇒tan θ=tan α=((a^2 −b^2 )/(2ab)) ✓
$$\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\mathrm{sin}\:\theta+\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\mathrm{cos}\:\theta=\mathrm{1} \\ $$$${since}\:\left(\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{1}, \\ $$$$\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{sin}\:\alpha,\:\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{cos}\:\alpha \\ $$$$\mathrm{sin}\:\alpha\:\mathrm{sin}\:\theta+\mathrm{cos}\:\alpha\:\mathrm{cos}\:\theta=\mathrm{1} \\ $$$$\mathrm{cos}\:\left(\theta−\alpha\right)=\mathrm{1} \\ $$$$\Rightarrow\theta−\alpha=\mathrm{2}{k}\pi \\ $$$$\Rightarrow\theta=\mathrm{2}{k}\pi+\alpha \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\mathrm{tan}\:\alpha=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ab}}\:\checkmark \\ $$
Answered by mr W last updated on 21/Jun/25
let t=tan (θ/2) (((a^2 −b^2 )2t)/(1+t^2 ))+((2ab(1−t^2 ))/(1+t^2 ))=a^2 +b^2 (a^2 +b^2 +2ab)t^2 −2(a^2 −b^2 )t+a^2 +b^2 −2ab=0 (a+b)^2 t^2 −2(a+b)(a−b)t+(a−b)^2 =0 [(a+b)t−(a−b)]^2 =0 ⇒t=((a−b)/(a+b)) tan θ=((2t)/(1−t^2 ))=((a^2 −b^2 )/(2ab)) ✓
$${let}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{2}{ab}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\right){t}^{\mathrm{2}} −\mathrm{2}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){t}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}=\mathrm{0} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} {t}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}\right)\left({a}−{b}\right){t}+\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left[\left({a}+{b}\right){t}−\left({a}−{b}\right)\right]^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{{a}−{b}}{{a}+{b}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ab}}\:\checkmark \\ $$
Commented by fantastic last updated on 21/Jun/25
Thank you very much
$${Thank}\:{you}\:{very}\:{much} \\ $$

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