CUET UG 2024 — Mathematics Applied-Mathematics

Starting from:

$$\sin y = x \sin(a + y)$$siny=xsin(a+y)

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

$$x = \dfrac{\sin y}{\sin(a + y)}$$x=sin(a+y)siny​

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

---

Differentiating $x$x with respect to $y$y using the quotient rule:

$$\dfrac{dx}{dy} = \dfrac{\sin(a+y) \cdot \cos y - \sin y \cdot \cos(a+y)}{\sin^2(a+y)}$$dydx​=sin2(a+y)sin(a+y)⋅cosy−siny⋅cos(a+y)​

---

The numerator matches the sine subtraction identity: $\sin A \cos B - \cos A \sin B = \sin(A - B)$sinAcosB−cosAsinB=sin(A−B)

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

$$\sin(a+y)\cos y - \sin y \cos(a+y) = \sin((a+y) - y) = \sin a$$sin(a+y)cosy−sinycos(a+y)=sin((a+y)−y)=sina

So:

$$\dfrac{dx}{dy} = \dfrac{\sin a}{\sin^2(a+y)}$$dydx​=sin2(a+y)sina​

---

Flipping to get $\dfrac{dy}{dx}$dxdy​:

$$\dfrac{dy}{dx} = \dfrac{1}{\frac{dx}{dy}} = \dfrac{\sin^2(a+y)}{\sin a}$$dxdy​=dydx​1​=sinasin2(a+y)​

---

$$\boxed{\dfrac{dy}{dx} = \dfrac{\sin^2(a+y)}{\sin a}}$$dxdy​=sinasin2(a+y)​​