If $\mathrm{x}=\sqrt{{\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}}\text{ and }\mathrm{y}=\sqrt{{\mathrm{a}}^{{\cos }^{-1} \mathrm{t}}}$then show that $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{\mathrm{x}}.$

$\mathrm{Given}, \mathrm{x}=\sqrt{{\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}}\text{ and }\mathrm{y}=\sqrt{{\mathrm{a}}^{{\cos }^{-1} \mathrm{t}}}$
$\mathrm{Consider} \mathrm{x}={\left({\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}\right)}^{1/2}$
On differentiating both sides w.r.t.  t, we get
$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{2}{\left({\mathrm{a}}^{\sin -1}\mathrm{t}\right)}^{-1/2}\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}\right)$
$\text{ [by using chain rule of derivative] }$
$\begin{array}{l}=\frac{1}{2}{\left({\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}\right)}^{-1/2}{\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}\log \mathrm{a}\frac{\mathrm{d}}{\mathrm{dt}}\left({\sin }^{-1} \mathrm{t}\right)\\ \left[\because \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{a}}^{\mathrm{x}}\right)={\mathrm{a}}^{\mathrm{x}}\log \mathrm{a}\right]\\ =\frac{1}{2}{\left({\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}\right)}^{-1/2}{\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}\log \mathrm{a}\cdot \frac{1}{\sqrt{1-{\mathrm{t}}^2}}\\ =\frac{1}{2}{\left({\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}\right)}^{1/2}\log \mathrm{a}\cdot \frac{1}{\sqrt{1-{\mathrm{t}}^2}}\end{array}$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\frac{1}{2}\sqrt{{\mathrm{a}}^{{\sin }^{-1}\mathrm{t}}}\cdot \log \mathrm{a}}{\sqrt{1-{\mathrm{t}}^2}}........\left(\mathrm{i}\right)$
Now, consider $\mathrm{y}={\left({\mathrm{a}}^{{\cos }^{-1} \mathrm{t}}\right)}^{1/2}$
On differentiating both sides w.r.t. t, we
$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{1}{2}{\left({\mathrm{a}}^{{\cos }^{-1} \mathrm{t}}\right)}^{-1/2}\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{a}}^{{\cos }^{-1} \mathrm{t}}\right)$
[by using chain rule of derivative]
$\begin{array}{l}=\frac{1}{2}{\left({\mathrm{a}}^{{\cos }^{-1} \mathrm{t}}\right)}^{-1/2}{\mathrm{a}}^{{\cos }^{-1} \mathrm{t}}\log \mathrm{a}\frac{\mathrm{d}}{\mathrm{dt}}\left({\cos }^{-1} \mathrm{t}\right)\\ \left[\because \frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{a}}^{\mathrm{x}}\right)={\mathrm{a}}^{\mathrm{x}}\log \mathrm{a}\right]\\ =\frac{1}{2}\left({\mathrm{a}}^{{\cos }^{-1}}{\mathrm{t}}^{1/2}\log \mathrm{a}\cdot \frac{(-1)}{\sqrt{1-{\mathrm{t}}^2}}\right.\\ \end{array}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{-\frac{1}{2}\sqrt{{\mathrm{a}}^{{\cos }^{-1} \mathrm{t}}}\cdot \log \mathrm{a}}{\sqrt{1-{\mathrm{t}}^2}} .....\left(\mathrm{ii}\right)$
On dividing Eq. (ii) by Eq. (i), we get
$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}=\frac{\left.\frac{-\frac{1}{2}\sqrt{{\mathrm{a}}^{{\cos }^{-1} \mathrm{t}}}\log \mathrm{a}}{\sqrt{1-{\mathrm{t}}^2}}\right)}{\left.\frac{\frac{1}{2}\sqrt{{\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}}\log \mathrm{a}}{\sqrt{1-{\mathrm{t}}^2}}\right)}\\ =-\frac{\sqrt{{\mathrm{a}}^{{\cos }^{-1} \mathrm{t}}}}{\sqrt{{\mathrm{a}}^{{\sin }^{-1} \mathrm{t}}}}=-\frac{\mathrm{y}}{\mathrm{x}}\\ \end{array}$