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Question Number 121633 by liberty last updated on 10/Nov/20

$$\mathrm{Given}\mathrm{implicit}\mathrm{expression}\{\begin{array}{l}\mathrm{x}=\tan \alpha \\ \mathrm{y}=\tan \beta \end{array}$$$$\mathrm{where}\mathrm{x},\mathrm{y},\alpha \mathrm{and}\beta \mathrm{is}\mathrm{variable}.\mathrm{Prove}\mathrm{that}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1+{\mathrm{y}}^2}{1+{\mathrm{x}}^2}=0.$$

Commented by Dwaipayan Shikari last updated on 10/Nov/20

$$x=tan\alpha \Rightarrow {tan}^{-1}x=\alpha \Rightarrow \frac{1}{1+x^2}=\frac{d\alpha }{dx}$$$$y=tan\beta \Rightarrow {tan}^{-1}y=\beta \Rightarrow \frac{1}{1+y^2}=\frac{d\beta }{dy}$$$$\frac{dy}{dx}.\frac{d\alpha }{d\beta }=\frac{1+y^2}{1+x^2}$$

Commented by liberty last updated on 10/Nov/20

$$\mathrm{not}\mathrm{proved}?$$

Commented by Dwaipayan Shikari last updated on 10/Nov/20

$$Wehavetoknowrelationbetween\alpha and\beta$$

Commented by Dwaipayan Shikari last updated on 10/Nov/20

$$If\alpha +\beta =0then$$$$-\frac{dy}{dx}=\frac{1+y^2}{1+x^2}\Rightarrow \frac{dy}{dx}+\frac{1+y^2}{1+x^2}=0$$

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