If $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2 t}{1+t^2}$, then $\frac{d y}{d x}$ is equal to
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If $x=\frac{1-t^2}{1+t^2}$ and $y=\frac{2 t}{1+t^2}$, then $\frac{d y}{d x}$ is equal to

(A) $-\frac{y}{x}$

(B) $\frac{y}{x}$

(C) $-\frac{x}{y}$

(D) $\frac{x}{y}$

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Best answer

SOLUTION —

Put $t=\tan \theta$ in both the equations, we get

$x=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}=\cos 2 \theta \text { and } y=\frac{2 \tan \theta}{1+\tan ^2 \theta}=\sin 2 \theta$

$\Rightarrow \quad \frac{d x}{d \theta}=-2 \sin 2 \theta$

and, $\frac{d y}{d \theta}=2 \cos 2 \theta$

$\therefore \quad \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=-\frac{\cos 2 \theta}{\sin 2 \theta}=-\frac{x}{y}$

So, The correct option will be (C).

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