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Question Number 206500 by MATHEMATICSAM last updated on 16/Apr/24
If asin^2 θ + bcos^2 θ = c, bsin^2 φ + acos^2 φ = d and atanθ = btanφ then prove that (1/a) + (1/b) = (1/c) + (1/d) .
$$\mathrm{If}\:{a}\mathrm{sin}^{\mathrm{2}} \theta\:+\:{b}\mathrm{cos}^{\mathrm{2}} \theta\:=\:{c},\:{b}\mathrm{sin}^{\mathrm{2}} \phi\:+\:{a}\mathrm{cos}^{\mathrm{2}} \phi\:=\:{d} \\ $$$$\mathrm{and}\:{a}\mathrm{tan}\theta\:=\:{b}\mathrm{tan}\phi\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:=\:\frac{\mathrm{1}}{{c}}\:+\:\frac{\mathrm{1}}{{d}}\:. \\ $$
Answered by Berbere last updated on 17/Apr/24
suppse cos(θ).cos(φ)≠0 (1)⇔atan^2 (θ)+b=c(1+tan^2 (θ))⇔tan^2 (θ)=((c−b)/(a−c)) (2)⇔((d−a)/(b−d))=tan^2 (φ) (b^2 /a^2 )=(((c−b)(b−d))/((a−c)(d−a)))⇔(1−(c/a))((d/a)−1)=((c/b)−1)(1−(d/b)) ⇔((c+d)/a)−((cd)/a^2 )=((c+d)/b)−((cd)/b^2 ) ⇔(c+d)((1/a)−(1/b))+cd((1/b^2 )−(1/a^2 ))=0 ⇔(c+d)((1/a)−(1/b))−cd((1/a)−(1/b))((1/a)+(1/b))=0 ⇔((1/a)−(1/b))(c+d−cd((1/a)+(1/b)))=0 if a=b⇒c=d=a=b⇒True if c+d−cd((1/a)+(1/b))=0⇒cd((1/a)+(1/b))=c+d if cd≠0≪The result implie abcd#0 ⇒(1/a)+(1/b)=(1/d)+(1/c) case (2) cos(θ)cos(φ)=0 impossibl due to the condition (atan (θ)=btan (φ)) so abcd#0;cos(θ)cos(φ)#0 (1/a)+(1/b)=(1/d)+(1/c)
$${suppse}\:{cos}\left(\theta\right).{cos}\left(\phi\right)\neq\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow{a}\mathrm{tan}\:^{\mathrm{2}} \left(\theta\right)+{b}={c}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left(\theta\right)\right)\Leftrightarrow{tan}^{\mathrm{2}} \left(\theta\right)=\frac{{c}−{b}}{{a}−{c}} \\ $$$$\left(\mathrm{2}\right)\Leftrightarrow\frac{{d}−{a}}{{b}−{d}}=\mathrm{tan}\:^{\mathrm{2}} \left(\phi\right) \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\frac{\left({c}−{b}\right)\left({b}−{d}\right)}{\left({a}−{c}\right)\left({d}−{a}\right)}\Leftrightarrow\left(\mathrm{1}−\frac{{c}}{{a}}\right)\left(\frac{{d}}{{a}}−\mathrm{1}\right)=\left(\frac{{c}}{{b}}−\mathrm{1}\right)\left(\mathrm{1}−\frac{{d}}{{b}}\right) \\ $$$$\Leftrightarrow\frac{{c}+{d}}{{a}}−\frac{{cd}}{{a}^{\mathrm{2}} }=\frac{{c}+{d}}{{b}}−\frac{{cd}}{{b}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\left({c}+{d}\right)\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right)+{cd}\left(\frac{\mathrm{1}}{{b}^{\mathrm{2}} }−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\Leftrightarrow\left({c}+{d}\right)\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right)−{cd}\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}}\right)\left({c}+{d}−{cd}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\right)=\mathrm{0} \\ $$$${if}\:{a}={b}\Rightarrow{c}={d}={a}={b}\Rightarrow{True} \\ $$$${if}\:{c}+{d}−{cd}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)=\mathrm{0}\Rightarrow{cd}\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)={c}+{d} \\ $$$${if}\:{cd}\neq\mathrm{0}\ll{The}\:{result}\:{implie}\:{abcd}#\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{d}}+\frac{\mathrm{1}}{{c}} \\ $$$${case}\:\left(\mathrm{2}\right)\:\:{cos}\left(\theta\right){cos}\left(\phi\right)=\mathrm{0}\:{impossibl}\:{due}\:{to}\:{the}\:{condition} \\ $$$$\left({a}\mathrm{tan}\:\left(\theta\right)={b}\mathrm{tan}\:\left(\phi\right)\right) \\ $$$${so}\:{abcd}#\mathrm{0};{cos}\left(\theta\right){cos}\left(\phi\right)#\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{d}}+\frac{\mathrm{1}}{{c}} \\ $$$$ \\ $$
Answered by Frix last updated on 17/Apr/24
One possible path: cos^2 θ =((a−c)/(a−b))∧cos^2 φ =−((b−d)/(a−b)) atan θ =btan φ =x ⇒ θ=tan^(−1) (x/a) ∧φ=tan^(−1) (x/b) ⇒ x^2 =−((a^2 (b−c))/(a−c))=−(((a−d)b^2 )/(b−d)) ⇒ d=−((abc)/(ab−ac−bd)) ⇒ (1/d)=(1/a)+(1/b)−(1/c) (1/a)+(1/b)=(1/c)+(1/d)
$$\mathrm{One}\:\mathrm{possible}\:\mathrm{path}: \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\theta\:=\frac{{a}−{c}}{{a}−{b}}\wedge\mathrm{cos}^{\mathrm{2}} \:\phi\:=−\frac{{b}−{d}}{{a}−{b}} \\ $$$${a}\mathrm{tan}\:\theta\:={b}\mathrm{tan}\:\phi\:={x}\:\Rightarrow \\ $$$$\theta=\mathrm{tan}^{−\mathrm{1}} \:\frac{{x}}{{a}}\:\wedge\phi=\mathrm{tan}^{−\mathrm{1}} \:\frac{{x}}{{b}} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} =−\frac{{a}^{\mathrm{2}} \left({b}−{c}\right)}{{a}−{c}}=−\frac{\left({a}−{d}\right){b}^{\mathrm{2}} }{{b}−{d}} \\ $$$$\Rightarrow \\ $$$${d}=−\frac{{abc}}{{ab}−{ac}−{bd}} \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{1}}{{d}}=\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{c}} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{c}}+\frac{\mathrm{1}}{{d}} \\ $$

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