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Question Number 32971 by math1967 last updated on 08/Apr/18
If siny=xsin(a+y) show that (dy/dx)=((sina)/(1−2xcosa+x^2 ))
$${If}\:{siny}={xsin}\left({a}+{y}\right)\:{show}\:{that} \\ $$$$\frac{{dy}}{{dx}}=\frac{{sina}}{\mathrm{1}−\mathrm{2}{xcosa}+{x}^{\mathrm{2}} } \\ $$$$ \\ $$
Answered by math1967 last updated on 06/May/18
siny=xsinacosy +xcosasiny ⇒siny(1−xcosa)=xsinacosy ⇒tany=((xsina)/((1−xcosa))) ⇒sec^2 y(dy/dx) =((sina)/((1−xcosa)^2 )) ⇒(dy/dx)=((sina)/((1−xcosa)^2 ))×(1/(1+tan^2 y)) (dy/dx)=((sina)/((1−xcosa)^2 ))×(1/(1+((x^2 sin^2 a)/((1−xcosa)^2 )))) ⇒(dy/dx)=((sina)/((1−xcosa)^2 ))×(((1−xcosa)^2 )/(1−2xcosa+x^2 (sin^2 a+cos^2 a))) =((sina)/(1−2xcosa+x^2 ))
$${siny}={xsinacosy}\:+{xcosasiny} \\ $$$$\Rightarrow{siny}\left(\mathrm{1}−{xcosa}\right)={xsinacosy} \\ $$$$\Rightarrow{tany}=\frac{{xsina}}{\left(\mathrm{1}−{xcosa}\right)} \\ $$$$\Rightarrow{sec}^{\mathrm{2}} {y}\frac{{dy}}{{dx}}\:=\frac{{sina}}{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{sina}}{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {y}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{sina}}{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+\frac{{x}^{\mathrm{2}} {sin}^{\mathrm{2}} {a}}{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{sina}}{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} }×\frac{\left(\mathrm{1}−{xcosa}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{xcosa}+{x}^{\mathrm{2}} \left({sin}^{\mathrm{2}} {a}+{cos}^{\mathrm{2}} {a}\right)} \\ $$$$=\frac{{sina}}{\mathrm{1}−\mathrm{2}{xcosa}+{x}^{\mathrm{2}} } \\ $$

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