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  • If <img alt="x=\frac{1-t^2}{1+t^2} \text { and } y=\frac{2 t}{1+t^2} \text {, then } \frac{d y}{d x}=" src="https://entrancecorner.oncodecogs.com/gif.latex?x%3D%5Cfrac%7B1-t%5E2%7D%7B1&plu

If x=\frac{1-t^2}{1+t^2} \text { and } y=\frac{2 t}{1+t^2} \text {, then } \frac{d y}{d x}=

Option: 1

-y/x


Option: 2

y/x


Option: 3

-x/y


Option: 4

x/y


Answers (1)

best_answer

x=\frac{1-t^2}{1+t^2} \text { and } y=\frac{2 t}{1+t^2}

Put t=\tan \theta in both the equations, we get x=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}=\cos 2 \theta and y=\frac{2 \tan \theta}{1+\tan ^2 \theta}=\sin 2 \theta

Differentiating both the equations, we get 

\frac{d x}{d \theta}=-2 \sin 2 \theta and  \frac{d y}{d \theta}=2 \cos 2 \theta .

Therefore 

\frac{d y}{d x}=-\frac{\cos 2 \theta}{\sin 2 \theta}=-\frac{x}{y}

 

Posted by

Deependra Verma

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