Question

\(\text{ If } \sin y = x \sin(a + y), \text{ then } \frac{dy}{dx} \text{ is:}\)

• A. \( \frac{\sin \frac{a}{2}}{\sin(a + y)} \)
• B. \( \frac{\sin(a + y)}{\sin^2 a} \)
• C. \( \frac{\sin(a + y)}{\sin a} \)
• D. \( \frac{\sin^2(a + y)}{\sin a} \)

Answer 1

The Correct Option is D

The given equation is:

\(\sin y = x \sin(a + y)\).

Differentiate both sides with respect to \(x\):

\(\cos y \frac{dy}{dx} = \sin(a + y) + x \cos(a + y) \frac{dy}{dx}\).

Rearrange to isolate \(\frac{dy}{dx}\):

\(\frac{dy}{dx} (\cos y - x \cos(a + y)) = \sin(a + y)\).

Simplify:

\(\frac{dy}{dx} = \frac{\sin(a + y)}{\cos y - x \cos(a + y)}\).

From the original equation \(\sin y = x \sin(a + y)\), rewrite \(x\) as:

\(x = \frac{\sin y}{\sin(a + y)}\).

Substitute \(x\) into the denominator:

\(\cos y - x \cos(a + y) = \cos y - \frac{\sin y \cos(a + y)}{\sin(a + y)}\).

Simplify the denominator:

\(\cos y - x \cos(a + y) = \frac{\cos y \sin(a + y) - \sin y \cos(a + y)}{\sin(a + y)} = \frac{\sin a}{\sin(a + y)}\).

Substitute this back into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\sin(a + y)}{\frac{\sin a}{\sin(a + y)}} = \frac{\sin^2(a + y)}{\sin a}\).

Thus:

\(\boxed{\frac{\sin^2(a + y)}{\sin a}}\).