We have, $y=(\cos x)^{(\cos x)^{\left.(\cos x)^{-\infty}\right)}}$

$$\begin{array}{ll} \Rightarrow & y=(\cos x)^y \\ \therefore & \log y=\log (\cos x)^y \\ \Rightarrow & \log y=y \log \cos x \end{array}$$

$$\begin{aligned} &\text { On differentiating w.r.t. } x \text {, we get }\\ &\begin{aligned} & \frac{1}{y} \cdot \frac{d y}{d x}=y \cdot \frac{d}{d x} \log \cos x+\log \cos x \cdot \frac{d y}{d x} \\ \Rightarrow \quad & \frac{1}{y} \cdot \frac{d y}{d x}=\frac{y}{\cos x} \cdot \frac{d}{d x} \cos x+\log \cos x \cdot \frac{d y}{d x} \end{aligned} \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad \frac{d y}{d x}\left[\frac{1}{y}-\log \cos x\right] & =\frac{-y \sin x}{\cos x}=-y \tan x \\ \therefore \quad \frac{d y}{d x} & =\frac{-y^2 \tan x}{(1-y \log \cos x)} \\ & =\frac{y^2 \tan x}{y \log \cos x-1}\quad \text{Hence proved.} \end{aligned}$$

$$\begin{aligned} &\text { We have, }\\ &\begin{array}{cc} & x \sin (a+y)+\sin a \cdot \cos (a+y)=0 \\ \Rightarrow & x \sin (a+y)=-\sin a \cdot \cos (a+y) \\ \Rightarrow & x=\frac{-\sin a \cdot \cos (a+y)}{\sin (a+y)} \end{array} \end{aligned}$$

$\Rightarrow \quad x=-\sin a \cdot \cot (a+y)$

$\therefore \quad \frac{d x}{d y}=-\sin a \cdot\left[-\operatorname{cosec}^2(a+y)\right] \cdot \frac{d}{d y}(a+y)$

$$\begin{aligned} &\begin{aligned} & =\sin a \cdot \frac{1}{\sin ^2(a+y)} \cdot 1 \\ & =\frac{\sin ^2(a+y)}{\sin a} \end{aligned}\\ &\text { Hence proved. } \end{aligned}$$

We have, $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$

$\begin{aligned} & \text { On putting } x=\sin \alpha \text { and } y=\sin \beta \text {, we get } \\ & \qquad \sqrt{1-\sin ^2 \alpha}+\sqrt{1-\sin ^2 \beta}=a(\sin \alpha-\sin \beta)\end{aligned}$

$$\begin{array}{ll} \Rightarrow & \cos \alpha+\cos \beta=a(\sin \alpha-\sin \beta) \\ \Rightarrow & 2 \cos \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}=a\left(2 \cos \frac{\alpha+\beta}{2} \cdot \sin \frac{\alpha-\beta}{2}\right) \end{array}$$

$$\begin{array}{ll} \Rightarrow & \cos \frac{\alpha-\beta}{2}=a \sin \frac{\alpha-\beta}{2} \\ \Rightarrow & \cot \frac{\alpha-\beta}{2}=a \end{array}$$

$$\begin{array}{l} \Rightarrow & \frac{\alpha-\beta}{2} =\cot ^{-1} a \\ \Rightarrow & \alpha-\beta =2 \cot ^{-1} a \\ \Rightarrow & \sin ^{-1} x-\sin ^{-1} y =2 \cot ^{-1} a \quad[\because x=\sin \alpha \text { and } y=\sin \beta] \end{array}$$

$$\begin{aligned} &\text { On differentiating both sides w.r.t. } x \text {, we get }\\ &\begin{array}{ll} & \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}} \frac{d y}{d x}=0 \\ \therefore & \frac{d y}{d x}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}=\sqrt{\frac{1-y^2}{1-x^2}}\quad\text{Hence proved.} \end{array} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { We have, }\quad y & =\tan ^{-1} x \quad \text { [on differentiating w.r.t. } x \text { ] }\\ \therefore\quad \frac{d y}{d x} & =\frac{1}{1+x^2}\quad \text { [again differentiating w.r.t. } x \text { ] } \end{aligned} \end{aligned}$$

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

$$\begin{aligned} & =-1\left(1+x^2\right)^{-2} \cdot \frac{d}{d x}\left(1+x^2\right) \\ & =-\frac{1}{\left(1+x^2\right)^2} \cdot 2 x \\ & =\frac{-2 \tan y}{\left(1+\tan ^2 y\right)^2} \quad\left[\because y=\tan ^{-1} x \Rightarrow \tan y=x\right] \end{aligned}$$

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

We have, $f(x)=x(x-1)^2$ in $[0,1]$.

(i) Since, $f(x)=x(x-1)^2$ is a polynomial function.

So, it is continuous in $[0,1]$.

$$\begin{aligned} &\text { (ii) Now, }\\ &\begin{aligned} f^{\prime}(x) & =x \cdot \frac{d}{d x}(x-1)^2+(x-1)^2 \frac{d}{d x} x \\ & =x \cdot 2(x-1) \cdot 1+(x-1)^2 \\ & =2 x^2-2 x+x^2+1-2 x \\ & =3 x^2-4 x+1 \text { which exists in }(0,1) \end{aligned} \end{aligned}$$

So, $f(x)$ is differentiable in $(0,1)$. (iii) Now, $f(0)=0$ and $f(1)=0 \Rightarrow f(0)=f(1)$

$f$ satisfies the above conditions of Rolle's theorem.

Hence, by Rolle's theorem $\exists c \in(0,1)$ such that

$$\begin{aligned} f^{\prime}(c) & =0 \\ \Rightarrow\quad 3 c^2-4 c+1 & =0 \\ \Rightarrow\quad 3 c^2-3 c-c+1 & =0 \\ \Rightarrow\quad 3 c(c-1)-1(c-1) & =0 \\ \Rightarrow\quad (3 c-1)(c-1) & =0 \\ \Rightarrow\quad c & =\frac{1}{3}, 1 \Rightarrow \frac{1}{3} \in(0,1) \end{aligned}$$

Thus, we see that there exists a real number c in the open interval $(0,1)$. Hence, Rolle's theorem has been verified.