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Question Number 211019 by peter frank last updated on 26/Aug/24

Answered by som(math1967) last updated on 26/Aug/24

let x=sinα y=sinβ cosα+cosβ=a(sinα−sinβ) ⇒((2cos((α+β)/2)cos((α−β)/2))/(2sin((α−β)/2)cos((α+β)/2)))=a ⇒cot((α−β)/2)=a ⇒(1/2)(α−β)=cot^(−1) a ⇒ sin^(−1) x−sin^(−1) y=2cot^(−1) a ⇒ (1/( (√(1−x^2 )))) −(1/( (√(1−y^2 ))))×(dy/dx)=0 ∴ (dy/dx)=(√((1−y^2 )/(1−x^2 )))

$$\:{let}\:{x}={sin}\alpha\:\:{y}={sin}\beta \\ $$$$\:{cos}\alpha+{cos}\beta={a}\left({sin}\alpha−{sin}\beta\right) \\ $$$$\Rightarrow\frac{\mathrm{2}{cos}\frac{\alpha+\beta}{\mathrm{2}}{cos}\frac{\alpha−\beta}{\mathrm{2}}}{\mathrm{2}{sin}\frac{\alpha−\beta}{\mathrm{2}}{cos}\frac{\alpha+\beta}{\mathrm{2}}}={a} \\ $$$$\Rightarrow{cot}\frac{\alpha−\beta}{\mathrm{2}}={a} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left(\alpha−\beta\right)=\mathrm{cot}^{−\mathrm{1}} {a} \\ $$$$\Rightarrow\:\mathrm{sin}^{−\mathrm{1}} {x}−\mathrm{sin}^{−\mathrm{1}} {y}=\mathrm{2cot}^{−\mathrm{1}} {a} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}×\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\:\therefore\:\frac{{dy}}{{dx}}=\sqrt{\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$

Commented by peter frank last updated on 27/Aug/24

thank you som

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{som} \\ $$

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