if cos y is equal to x cos (a + y) with cos a not equal to plus minus one, prove that dy/dx is equal to cos square (a + y) whole upon sin a i.e. dy/dx=(cos^2(a+y))/sin a
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\[ \begin{align*} \cos(y)&=x\cos(a+y)\implies x=\frac{\cos(y)}{\cos(a+y)}\\ \frac{d}{dx}\cos(y)&=\frac{d}{dx}\left(x\cos(a+y)\right)\\ -\sin(y)\frac{dy}{dx}&=\cos(a+y)-x\sin(a+y)\frac{dy}{dx}\\ \frac{\cos(y)}{\cos(a+y)}\sin(a+y)\frac{dy}{dx}-\sin(y)&=\cos(a+y)\\ \frac{dy}{dx}\left(\sin(a+y)\cos(y)-\sin(y)\cos(a+y)\right)&=\cos^2(a+y)\\ \frac{dy}{dx}\sin(a)&=\cos^2(a+y)\\ \frac{dy}{dx}&=\frac{\cos^2(a+y)}{\sin(a)} \end{align*} \]
thank you thomas, in the 4th line(step) i think there should be dy/dx with sin(y). will you please explain it. after fourth step it is not clear.
Yes I missed dy/dx in the fourth line. Thanks for pointing that out.
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