Question

If sin y = sin 3t and x = sin t, then \(\frac{dy}{dx}\) =

• A. \(\frac{3}{\sqrt4-x^2}\)
• B. \(\frac{3}{\sqrt1-x^2}\)
• C. \(\frac{1}{\sqrt4-x^2}\)
• D. \(\frac{-1}{\sqrt4 - x^2}\)

Answer 1

The Correct Option is B

To solve this problem, we need to find \( \frac{dy}{dx} \), given that \( \sin y = \sin 3t \) and \( x = \sin t \).

1. From \( \sin y = \sin 3t \), taking principal value, \( y = 3t \).

2. Differentiating with respect to \( t \):
\( \frac{dy}{dt} = 3 \).

3. Given \( x = \sin t \), so
\( \frac{dx}{dt} = \cos t \).

4. Using chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3}{\cos t} \]

5. Since \( \sin t = x \), we have
\[ \cos t = \sqrt{1 - \sin^2 t} = \sqrt{1 - x^2} \]

6. Therefore,
\[ \frac{dy}{dx} = \frac{3}{\sqrt{1 - x^2}} \]

Thus, the required answer is \( \frac{3}{\sqrt{1 - x^2}} \).