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If sin⁡y=xsin⁡(a+y)\sin y=x \sin (a+y), then dydx\frac{d y}{d x} is :

Solution

✅ Correct Option: 4

Starting from:

sin⁡y=xsin⁡(a+y)\sin y = x \sin(a + y)

Let's solve for xx first, as this makes differentiation much easier:

x=sin⁡ysin⁡(a+y)x = \dfrac{\sin y}{\sin(a + y)}

We'll find dxdy\dfrac{dx}{dy} and then flip it to get dydx=1dxdy\dfrac{dy}{dx} = \dfrac{1}{\frac{dx}{dy}}.


Differentiating xx with respect to yy using the quotient rule:

dxdy=sin⁡(a+y)⋅cos⁡y−sin⁡y⋅cos⁡(a+y)sin⁡2(a+y)\dfrac{dx}{dy} = \dfrac{\sin(a+y) \cdot \cos y - \sin y \cdot \cos(a+y)}{\sin^2(a+y)}


The numerator matches the sine subtraction identity: sin⁡Acos⁡B−cos⁡Asin⁡B=sin⁡(A−B)\sin A \cos B - \cos A \sin B = \sin(A - B)

With A=(a+y)A = (a+y) and B=yB = y:

sin⁡(a+y)cos⁡y−sin⁡ycos⁡(a+y)=sin⁡((a+y)−y)=sin⁡a\sin(a+y)\cos y - \sin y \cos(a+y) = \sin((a+y) - y) = \sin a

So:

dxdy=sin⁡asin⁡2(a+y)\dfrac{dx}{dy} = \dfrac{\sin a}{\sin^2(a+y)}


Flipping to get dydx\dfrac{dy}{dx}:

dydx=1dxdy=sin⁡2(a+y)sin⁡a\dfrac{dy}{dx} = \dfrac{1}{\frac{dx}{dy}} = \dfrac{\sin^2(a+y)}{\sin a}


dydx=sin⁡2(a+y)sin⁡a\boxed{\dfrac{dy}{dx} = \dfrac{\sin^2(a+y)}{\sin a}}

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