Question

If \( x = \frac{1-t^2}{1+t^2} , y = \frac{2t}{1+t^2} \) then prove that \( \frac{dy}{dx} + \frac{x}{y} = 0 \).
Correct Answer: Proved

Hint:

Trigonometric substitutions like \( t = \tan \theta \) can significantly simplify parametric expressions involving forms like \( \frac{1-t^2}{1+t^2} \) and \( \frac{2t}{1+t^2} \).

Answer 1

Step 1: Understanding the Concept:
This is a parametric differentiation problem. We find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), then use \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Alternatively, we can use trigonometric substitution.
Step 2: Detailed Explanation:
Let \( t = \tan \theta \).
Then \( x = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \cos 2\theta \).
And \( y = \frac{2\tan \theta}{1+\tan^2 \theta} = \sin 2\theta \).
Now, we have:
\( \frac{dx}{d\theta} = -2 \sin 2\theta \)
\( \frac{dy}{d\theta} = 2 \cos 2\theta \)
Therefore, \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2 \cos 2\theta}{-2 \sin 2\theta} = -\frac{\cos 2\theta}{\sin 2\theta} \).
Since \( \cos 2\theta = x \) and \( \sin 2\theta = y \):
\[ \frac{dy}{dx} = -\frac{x}{y} \]
\[ \frac{dy}{dx} + \frac{x}{y} = 0 \]
Step 3: Final Answer:
The identity \( \frac{dy}{dx} + \frac{x}{y} = 0 \) is proven.