Question:
If \( |t|<1 \), \( \sin x = \frac{2t}{1+t^2} \), \( \tan y = \frac{2t}{1-t^2} \), then \( \frac{dy}{dx} \) is
If \( |t|<1 \), \( \sin x = \frac{2t}{1+t^2} \), \( \tan y = \frac{2t}{1-t^2} \), then \( \frac{dy}{dx} \) is
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Recognize standard parametrizations of double angle identities immediately.
Updated On: May 8, 2026
- \( \frac{1}{x} \)
- \( \frac{1}{2} \)
- \( -\frac{1}{2} \)
- \( -\frac{1}{x} \)
- \( 1 \)
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Solution and Explanation
Concept:
Recognize parametric forms:
\[
\sin x = \frac{2t}{1+t^2} = \sin(2\theta)
\]
Step 1: Identify
Let \( t = \tan \theta \)
Step 2: Then
\[ \sin x = \sin(2\theta) \Rightarrow x = 2\theta \]
Step 3: Also
\[ \tan y = \frac{2t}{1-t^2} = \tan(2\theta) \Rightarrow y = 2\theta \]
Step 4: Relation
\[ x = y \]
Step 5: Differentiate
\[ \frac{dy}{dx} = 1 \]
Step 1: Identify
Let \( t = \tan \theta \)
Step 2: Then
\[ \sin x = \sin(2\theta) \Rightarrow x = 2\theta \]
Step 3: Also
\[ \tan y = \frac{2t}{1-t^2} = \tan(2\theta) \Rightarrow y = 2\theta \]
Step 4: Relation
\[ x = y \]
Step 5: Differentiate
\[ \frac{dy}{dx} = 1 \]
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