$$\begin{array}{ll} \because & \sin x=\frac{2 t}{1+t^2} \quad\text{.... (i)}\\ \text { and } & \tan y=\frac{2 t}{1-t^2}\quad\text{.... (ii)} \end{array}$$

$$\begin{aligned} & \begin{aligned} \therefore \quad \frac{d}{d x} \sin x \cdot \frac{d x}{d t} & =\frac{d}{d t}\left(\frac{2 t}{1+t^2}\right) \\ \Rightarrow \quad \cos x \frac{d x}{d t} & =\frac{\left(1+t^2\right) \cdot \frac{d}{d t}(2 t)-(2 t) \cdot \frac{d}{d t}\left(1+t^2\right)}{\left(1+t^2\right)^2} \\ & =\frac{2\left(1+t^2\right)-2 t \cdot 2 t}{\left(1+t^2\right)^2}=\frac{2+2 t^2-4 t^2}{\left(1+t^2\right)^2} \\ \Rightarrow \quad \frac{d x}{d t} & =\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \cdot \frac{1}{\cos x} \\ \Rightarrow \quad \frac{d x}{d t} & =\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \cdot \frac{1}{\sqrt{1-\sin ^2 x}}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \cdot \frac{1}{\sqrt{1-\left(\frac{2 t}{\left.1+t^2\right)^2}\right.}} \end{aligned} \end{aligned}$$

$\Rightarrow \quad \frac{d x}{d t}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \cdot \frac{\left(1+t^2\right)}{\left(1-t^2\right)}=\frac{2}{1+t^2}\quad\text{..... (iii)}$

$$\begin{aligned} &\text { Also, }\\ &\begin{aligned} \frac{d}{d y} \tan y \cdot \frac{d y}{d t} & =\frac{d}{d t}\left(\frac{2 t}{1-t^2}\right) \\ \sec ^2 y \frac{d y}{d t} & =\frac{\left(1-t^2\right) \frac{d}{d t} \cdot(2 t)-2 t \cdot \frac{d}{d t}\left(1-t^2\right)}{\left(1-t^2\right)^2} \\ \frac{d y}{d t} & =\frac{2-2 t^2+4 t^2}{\left(1-t^2\right)^2} \cdot \frac{1}{\sec ^2 y} \\ & =\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2} \cdot \frac{1}{\left(1+\tan ^2 y\right)}=\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2} \cdot \frac{1}{1+\frac{4 t^2}{\left(1-t^2\right)^2}} \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2} \cdot \frac{\left(1-t^2\right)^2}{\left(1+t^2\right)^2}=\frac{2}{1+t^2} \quad\text{.... (iv)}\\ \therefore \quad \frac{d y}{d x} & =\frac{d y / d t}{d x / d t}=\frac{2 / 1+t^2}{2 / 1+t^2}=1\quad\text{[from Eqs. (iii) and (iv)]} \end{aligned}$$

$$\begin{aligned} & \because \quad x=\frac{1+\log t}{t^2} \text { and } y=\frac{3+2 \log t}{t} \\ & \therefore \quad \frac{d x}{d t}=\frac{t^2 \cdot \frac{d}{d t}(1+\log t)-(1+\log t) \cdot \frac{d}{d t} t^2}{\left(t^2\right)^2} \end{aligned}$$

$$\begin{aligned} & =\frac{t^2 \cdot \frac{1}{t}-(1+\log t) \cdot 2 t}{t^4}=\frac{t-(1+\log t) \cdot 2 t}{t^4} \\ & =\frac{t}{t^4}\left[1-2(1+\log t)=\frac{-1-2 \log t}{t^3}\right. \quad\text{.... (i)}\\ \text{and}\quad \frac{d y}{d t} & =\frac{t \cdot \frac{d}{d t}(3+2 \log t)-(3+2 \log t) \cdot \frac{d}{d t} t}{t^2} \\ & =\frac{t \cdot 2 \cdot \frac{1}{t}-(3+2 \log t) \cdot 1}{t^2} \\ & =\frac{2-3-2 \log t}{t^2}=\frac{-1-2 \log t}{t^2} \quad \quad\text{.... (ii)}\\ \therefore\quad \frac{d y}{d x} & =\frac{d y / d t}{d x / d t}=\frac{-1-2 \log t / t^2}{-1-2 \log t / t^3}=t \end{aligned}$$

$$\begin{aligned} & \because \quad x=e^{\cos 2 t} \text { and } y=e^{\sin 2 t} \\ & \therefore \quad \frac{d x}{d t}=\frac{d}{d t} e^{\cos 2 t}=e^{\cos 2 t} \cdot \frac{d}{d t} \cos 2 t \\ & =e^{\cos 2 t} \cdot(-\sin 2 t) \cdot \frac{d}{d t}(2 t) \end{aligned}$$

$$\begin{aligned} \frac{d x}{d t} & =-2 e^{\cos 2 t} \cdot \sin 2 t \quad\text{.... (i)}\\ \text{and}\quad \frac{d y}{d t} & =\frac{d}{d t} e^{\sin 2 t}=e^{\sin 2 t} \cdot \frac{d}{d t} \sin 2 t \\ & =e^{\sin 2 t} \cos 2 t \cdot \frac{d}{d t} 2 t \\ & =2 e^{\sin 2 t} \cdot \cos 2 t\quad\text{.... (ii)} \end{aligned}$$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d y / d t}{d x / d t}=\frac{2 e^{\sin 2 t} \cdot \cos 2 t}{-2 e^{\cos 2 t} \cdot \sin 2 t} \\ & =\frac{e^{\sin 2 t} \cdot \cos 2 t}{e^{\cos 2 t} \cdot \sin 2 t}\quad\text{.... (iii)} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { We know that, }\quad \log x & =\cos 2 t \cdot \log e=\cos 2 t \quad\text{.... (iv)}\\ \text { and }\quad \log y & =\sin 2 t \cdot \log e=\sin 2 t \quad\text{..... (v)}\\ \therefore\quad \frac{d y}{d x} & =\frac{-y \log x}{x \log y}\quad \text { [using Eqs. (iv) and (v) in Eq. (iii) and } x=e^{\cos 2 t}, y=e^{\sin 2 t} \text { ] } \end{aligned} \end{aligned}$$

Hence proved.

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

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

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

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

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

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

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

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

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

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

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