$$\begin{array}{rlrl} &\because & x & =a \sin 2 t(1+\cos 2 t) \text { and } y=b \cos 2 t(1-\cos 2 t) \\ & \therefore & \frac{d x}{d t} & =a\left[\sin 2 t \cdot \frac{d}{d t}(1+\cos 2 t)+(1+\cos 2 t) \cdot \frac{d}{d t} \sin 2 t\right] \end{array}$$

$$\begin{aligned} & =a\left[\sin 2 t \cdot(-\sin 2 t) \cdot \frac{d}{d t} 2 t+(1+\cos 2 t) \cdot \cos 2 t \cdot \frac{d}{d t} 2 t\right] \\ & =-2 a \sin ^2 2 t+2 a \cos 2 t(1+\cos 2 t) \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & \frac{d x}{d t}=-2 a\left[\sin ^2 2 t-\cos 2 t(1+\cos 2 t)\right] \quad\text{.... (i)}\\ \text { and } & \frac{d y}{d t}=b\left[\cos 2 t \cdot \frac{d}{d t}(1-\cos 2 t)+(1-\cos 2 t) \cdot \frac{d}{d t} \cos 2 t\right] \end{array}$$

$$\begin{aligned} & =b\left[\cos 2 t \cdot(\sin 2 t) \frac{d}{d t} 2 t+(1-\cos 2 t)(-\sin 2 t) \cdot \frac{d}{d t} 2 t\right] \\ & =b[2 \sin 2 t \cdot \cos 2 t+2(1-\cos 2 t)(-\sin 2 t)] \\ & =2 b[\sin 2 t \cdot \cos 2 t-(1-\cos 2 t) \sin 2 t]\quad\text{.... (ii)} \end{aligned}$$

$$\begin{array}{ll} \therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{-2 b[-\sin 2 t \cdot \cos 2 t+(1-\cos 2 t) \sin 2 t]}{-2 a\left[\sin ^2 2 t-\cos 2 t(1+\cos 2 t)\right]} \\ \Rightarrow \quad\left(\frac{d y}{d x}\right)_{t=\pi / 4}=\frac{b}{a} \frac{\left[-\sin \frac{\pi}{2} \cos \frac{\pi}{2}+\left(1-\cos \frac{\pi}{2}\right) \sin \frac{\pi}{2}\right]}{\left[\sin ^2 \frac{\pi}{2}-\cos \frac{\pi}{2}\left(1+\cos \frac{\pi}{2}\right)\right]} \end{array}$$

$$\begin{aligned} & =\frac{b}{a} \cdot \frac{(0+1)}{(1-0)} \quad \left[\because \sin \frac{\pi}{2}=1 \text { and } \cos \frac{\pi}{2}=0\right]\\ & =\frac{b}{a}\quad\text{Hence proved.} \end{aligned}$$

$$\begin{aligned} \because \quad & x =3 \sin t-\sin 3 t \text { and } y=3 \cos t-\cos 3 t \\ \therefore \quad & \frac{d x}{d t} =3 \cdot \frac{d}{d t} \sin t-\frac{d}{d t} \sin 3 t \end{aligned}$$

$=3 \cos t-\cos 3 t \cdot \frac{d}{d t} 3 t=3 \cos t-3 \cos 3 t\quad\text{.... (i)}$

$$\begin{aligned} &\begin{aligned} \text { and }\quad\frac{d y}{d t} & =3 \cdot \frac{d}{d t} \cos t-\frac{d}{d t} \cos 3 t \\ & =-3 \sin t+\sin 3 t \cdot \frac{d}{d t} 3 t \end{aligned} \end{aligned}$$

$$\begin{array}{ll} \frac{d y}{d t}=3 \sin 3 t-3 t \sin t \quad\text{.... (ii)}\\ \therefore \quad \frac{d y}{d x} =\frac{d y / d t}{d x / d t}=\frac{3(\sin 3 t-\sin t)}{3(\cos t-\cos 3 t)} \end{array}$$

$$\begin{aligned} \text { Now, } \quad \left(\frac{d y}{d x}\right)_{t=\pi / 3} & =\frac{\sin \frac{3 \pi}{3}-\sin \frac{\pi}{3}}{\left(\cos \frac{\pi}{3}-\cos 3 \frac{\pi}{3}\right)}=\frac{0-\sqrt{3} / 2}{\frac{1}{2}-(-1)} \\ & =\frac{-\sqrt{3} / 2}{3 / 2}=\frac{-\sqrt{3}}{3}=\frac{-1}{\sqrt{3}} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Let }\quad u & =\frac{x}{\sin x} \text { and } v=\sin x \\ \therefore\quad \frac{d u}{d x} & =\frac{\sin x \cdot \frac{d}{d x} x-x \cdot \frac{d}{d x} \sin x}{(\sin x)^2} \\ & =\frac{\sin x-x \cos x}{\sin ^2 x}\quad\text{.... (i)} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { and }\quad & \frac{d v}{d x}=\frac{d}{d x} \sin x=\cos x \quad\text{..... (ii)}\\ \therefore\quad & \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{\sin x-x \cos x / \sin ^2 x}{\cos x} \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\frac{\sin x-x \cos x}{\sin ^2 x \cos x}=\frac{\frac{\sin x-x \cos x}{\cos x}}{\frac{\sin ^2 x \cos x}{\cos x}} \quad\text{[dividing by cos x in both numerator and denominator]}\\ & =\frac{\tan x-x}{\sin ^2 x} \end{aligned}$$

Let $$u=\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x} \text { and } v=\tan ^{-1} x$$

$$\begin{array}{ll} \therefore & x=\tan \theta \\ \Rightarrow & u=\tan ^{-1} \frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta} \end{array}$$

$$\begin{aligned} & =\tan ^{-1} \frac{(\sec \theta-1) \cos \theta}{\sin \theta} \\ & =\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\ & =\tan ^{-1}\left[\frac{1-1+2 \sin ^2 \theta / 2}{2 \sin \theta / 2 \cdot \cos \theta / 2}\right] \quad\left[\because \cos \theta=1-2 \sin ^2 \theta\right] \end{aligned}$$

$$ \begin{aligned} & =\tan ^{-1}\left[\tan \frac{\theta}{2}\right] \\ & =\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x \\ \therefore\quad \frac{d u}{d x} & =\frac{1}{2} \frac{d}{d x} \tan ^{-1} x=\frac{1}{2} \cdot \frac{1}{1+x^2} \quad\text{.... (i)}\\ \text{and}\quad \frac{d v}{d x} & =\frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^2} \quad\text{.... (ii)}\\ \therefore\quad \frac{d u}{d v} & =\frac{d u / d x}{d v / d x} \\ & =\frac{1 / 2\left(1+x^2\right)}{1 /\left(1+x^2\right)}=\frac{\left(1+x^2\right)}{2\left(1+x^2\right)}=\frac{1}{2} \end{aligned}$$

We have, $\quad \sin (x y)+\frac{x}{y}=x^2-y$

On differentiating both sides w.r.t. $x$, we get $$\frac{d}{d x}(\sin x y)+\frac{d}{d x}\left(\frac{x}{y}\right)=\frac{d}{d x} x^2-\frac{d}{d x} y$$

$$\begin{array}{ll} \Rightarrow & \cos x y \cdot \frac{d}{d x}(x y)+\frac{y \frac{d}{d x} x-x \cdot \frac{d}{d x} y}{y^2}=2 x-\frac{d y}{d x} \\ \Rightarrow & \cos x y \cdot\left[x \cdot \frac{d}{d x} y+y \cdot \frac{d}{d x} \cdot x\right]+\frac{y-x \frac{d y}{d x}}{y^2}=2 x-\frac{d y}{d x} \\ \Rightarrow & x \cos x y \cdot \frac{d y}{d x}+y \cos x y+\frac{y}{y^2}-\frac{x}{y^2} \frac{d y}{d x}=2 x-\frac{d y}{d x} \end{array}$$

$$\begin{array}{ll} \Rightarrow & \frac{d y}{d x}\left[x \cos x y-\frac{x}{y^2}+1\right]=2 x-y \cos x y-\frac{y}{y^2} \\ \therefore & \frac{d y}{d x}=\left[\frac{2 x y-y^2 \cos x y-1}{y}\right]\left[\frac{y^2}{x y^2 \cos x y-x+y^2}\right] \end{array}$$

$=\frac{\left(2 x y-y^2 \cos x y-1\right) y}{\left(x y^2 \cos x y-x+y^2\right)}$