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Question Number 26446 by yesaditya22@gmail.com last updated on 25/Dec/17

Answered by ajfour last updated on 25/Dec/17

$$\cos \theta \sin \theta +\sin \theta \cos \theta =a$$$$\Rightarrow \sin 2\theta =a$$$$\frac{d\left(\sin \theta \right)}{d\left(\cos \theta \right)}=\frac{\cos \theta }{-\sin \theta }=-\sqrt{\frac{1-y^2}{1-x^2}}.$$

Commented by mrW1 last updated on 26/Dec/17

$$x=\sin \theta ,\sqrt{1-x^2}=\cos \theta$$$$y=\sin \alpha ,\sqrt{1-y^2}=\cos \alpha$$$$\cos \alpha \sin \theta +\sin \alpha \cos \theta =a$$$$\Rightarrow \sin (\theta +\alpha )=a$$$$\Rightarrow \theta +\alpha ={\sin }^{-1}a$$$$\Rightarrow \alpha ={\sin }^{-1}a-\theta$$$$\Rightarrow \frac{d\alpha }{d\theta }=-1$$$$\frac{dy}{dx}=\frac{dy}{d\alpha }\times \frac{1}{\frac{dx}{d\theta }}\times \frac{d\alpha }{d\theta }=\frac{\cos \alpha }{\cos \theta }\times (-1)=-\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$$

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