Let $$y=\tan ^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]$$

Put $$x^2=\cos 2 \theta$$

$$\begin{aligned} &\begin{aligned} \text { Put }\quad x^2 & =\cos 2 \theta \\ \therefore\quad y & =\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right) \end{aligned} \end{aligned}$$

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

$$\begin{aligned} & =\tan ^{-1} \tan \left(\frac{\pi}{4}+\theta\right) \quad\left[\because \tan (a+b)=\frac{\tan a+\tan b}{1-\tan a \cdot \tan b}\right] \\ & =\frac{\pi}{4}+\theta=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2 \quad\left[\because 2 \theta=\cos ^{-1} x^2 \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x^2\right] \\ \therefore \quad \frac{d y}{d x} & =\frac{d}{d x}\left(\frac{\pi}{4}\right)+\frac{d}{d x}\left(\frac{1}{2} \cos ^{-1} x^2\right) \\ & =0+\frac{1}{2} \cdot \frac{-1}{\sqrt{1-x^4}} \cdot \frac{d}{d x} x^2=\frac{1}{2} \cdot \frac{-2 x}{\sqrt{1-x^4}}=\frac{-x}{\sqrt{1-x^4}} \end{aligned}$$

$$\begin{array}{lll} \because & x=t+\frac{1}{t} \text { and } y=t-\frac{1}{t} \\ \therefore & \frac{d x}{d t}=\frac{d}{d t}\left(t+\frac{1}{t}\right) \quad \text { and } \quad \frac{d y}{d t}=\frac{d}{d t}\left(t-\frac{1}{t}\right) \\ \Rightarrow & \frac{d x}{d t}=1+(-1) t^{-2} \text { and } \quad \frac{d y}{d t}=1-(-1) t^{-2} \\ \Rightarrow & \frac{d x}{d t}=1-\frac{1}{t^2} \quad \text { and } \quad \frac{d y}{d t}=1+\frac{1}{t^2} \\ \Rightarrow & \frac{d x}{d t}=\frac{t^2-1}{t^2} \quad \text { and } \frac{d y}{d t}=\frac{t^2+1}{t^2} \\ \therefore & \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{t^2+1 / t^2}{t^2-1 / t^2}=\frac{t^2+1}{t^2-1} \end{array}$$

$\because\quad x=e^\theta\left(\theta+\frac{1}{\theta}\right)$ and $y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$

$\therefore \quad \frac{d x}{d \theta}=\frac{d}{d \theta}\left[e^\theta \cdot\left(\theta+\frac{1}{\theta}\right)\right]$

$$\begin{aligned} & =e^\theta \cdot \frac{d}{d \theta}\left(\theta+\frac{1}{\theta}\right)+\left(\theta+\frac{1}{\theta}\right) \cdot \frac{d}{d \theta} e^\theta \\ & =e^\theta\left(1-\frac{1}{\theta^2}\right)+\left(\theta+\frac{1}{\theta}\right) e^\theta \\ & =e^\theta\left(1-\frac{1}{\theta^2}+\theta+\frac{1}{\theta}\right) \\ & =e^\theta\left(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\right)\text{.... (i)} \end{aligned}$$

and $$\frac{d y}{d \theta}=\frac{d}{d \theta}\left[e^{-\theta} \cdot\left(\theta-\frac{1}{\theta}\right)\right]$$

$$\begin{aligned} & =e^{-\theta} \cdot \frac{d}{d \theta}\left(\theta-\frac{1}{\theta}\right)+\frac{d}{d \theta} e^{-\theta}\left(\theta-\frac{1}{\theta}\right) \\ & =e^{-\theta}\left(1+\frac{1}{\theta^2}\right)+\left(\theta-\frac{1}{\theta}\right) e^{-\theta} \cdot \frac{d}{d \theta}(-\theta) \\ & =e^{-\theta}\left[\frac{\theta^2+1}{\theta^2}-\frac{\theta^2-1}{\theta}\right]=e^{-\theta}\left[\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\right]\quad\text{.... (ii)} \end{aligned}$$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d y / d \theta}{d x / d \theta}=\frac{e^{-\theta}\left(\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\right)}{e^\theta\left(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\right)} \\ & =e^{-2 \theta}\left(\frac{-\theta^3+\theta^2+\theta+1}{\theta^3+\theta^2+\theta-1}\right) \end{aligned}$$

$$\begin{aligned} \because\quad x & =3 \cos \theta-2 \cos ^3 \theta \text { and } y=3 \sin \theta-2 \sin ^3 \theta \\ \therefore\quad \frac{d x}{d \theta} & =\frac{d}{d \theta}(3 \cos \theta)-\frac{d}{d \theta}\left(2 \cos ^3 \theta\right) \\ & =3 \cdot(-\sin \theta)-2 \cdot 3 \cos ^2 \theta \cdot \frac{d}{d \theta} \cdot \cos \theta \\ & =-3 \sin \theta+6 \cos ^2 \theta \sin \theta \\ \text{and}\quad\frac{d y}{d \theta} & =3 \cos A-2 \cdot 3 \sin ^2 \theta \cdot \frac{d}{d \theta} \cdot \sin \theta \\ & =3 \cos \theta-6 \sin ^2 \theta \cdot \cos \theta \\ \text{Now,}\quad \frac{d y}{d x} & =\frac{d y / d \theta}{d x / d \theta}=\frac{3 \cos \theta-6 \sin ^2 \theta \cos \theta}{-3 \sin \theta+6 \cos ^2 \theta \sin \theta} \\ & =\frac{3 \cos \theta\left(1-2 \sin ^2 \theta\right)}{3 \sin \theta\left(-1+2 \cos ^2 \theta\right)}=\cot \theta \cdot \frac{\cos 2 \theta}{\cos 2 \theta}=\cot \theta \end{aligned}$$

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

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

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

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

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