Question:
If \( x\sin(a+y) + \sin a \cos(a+y)=0 \), then \( \frac{dy}{dx} \) is equal to
If \( x\sin(a+y) + \sin a \cos(a+y)=0 \), then \( \frac{dy}{dx} \) is equal to
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Use given equation again after differentiation to simplify.
Updated On: Apr 23, 2026
- $\frac{\sin^2(a+y)}{\sin a}$
- $\frac{\cos^2(a+y)}{\cos a}$
- $\frac{\sin^2(a+y)}{\cos a}$
- $\frac{\cos^2(a+y)}{\sin a}$
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The Correct Option is A
Solution and Explanation
Concept:
Implicit differentiation.
Step 1: Differentiate both sides.
\[ \frac{d}{dx}[x\sin(a+y)] + \frac{d}{dx}[\sin a \cos(a+y)] = 0 \]
Step 2: Apply product rule.
\[ \sin(a+y) + x\cos(a+y)\frac{dy}{dx} - \sin a \sin(a+y)\frac{dy}{dx} = 0 \]
Step 3: Group $\frac{dy}{dx}$.
\[ \sin(a+y) + \frac{dy}{dx}[x\cos(a+y) - \sin a \sin(a+y)] = 0 \]
Step 4: Solve for $\frac{dy}{dx}$.
\[ \frac{dy}{dx} = \frac{-\sin(a+y)}{x\cos(a+y) - \sin a \sin(a+y)} \]
Step 5: Use original equation.
From given: \[ x\sin(a+y) = -\sin a \cos(a+y) \] Substitute and simplify: \[ \frac{dy}{dx} = \frac{\sin^2(a+y)}{\sin a} \] Conclusion:
Answer = Option (A)
Step 1: Differentiate both sides.
\[ \frac{d}{dx}[x\sin(a+y)] + \frac{d}{dx}[\sin a \cos(a+y)] = 0 \]
Step 2: Apply product rule.
\[ \sin(a+y) + x\cos(a+y)\frac{dy}{dx} - \sin a \sin(a+y)\frac{dy}{dx} = 0 \]
Step 3: Group $\frac{dy}{dx}$.
\[ \sin(a+y) + \frac{dy}{dx}[x\cos(a+y) - \sin a \sin(a+y)] = 0 \]
Step 4: Solve for $\frac{dy}{dx}$.
\[ \frac{dy}{dx} = \frac{-\sin(a+y)}{x\cos(a+y) - \sin a \sin(a+y)} \]
Step 5: Use original equation.
From given: \[ x\sin(a+y) = -\sin a \cos(a+y) \] Substitute and simplify: \[ \frac{dy}{dx} = \frac{\sin^2(a+y)}{\sin a} \] Conclusion:
Answer = Option (A)
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