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Answer:

$\frac{d y}{d x}=1$

Hint:

Use chain rule, product rule and  $\frac{d y}{d x}=\frac{\frac{d x}{d t}}{\frac{d x}{d t}}$

Given:

$x=\sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right), y=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right) ;-1

Solution:

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

\begin{aligned} & &\frac{d x}{d t}=\frac{d \sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)}{d t} \end{aligned}

$=\frac{d \sin ^{-1}\left(\frac{2 t}{1+t^{2}}\right)}{d\left(\frac{2 t}{1+t^{2}}\right)} \times \frac{d\left(\frac{2 t}{1+t^{2}}\right)}{d t}$                                                             [Using chain rule]

$=\frac{1}{\sqrt{1-\left(\frac{2 t}{1+t^{2}}\right)^{2}}} \times \frac{\left(1+t^{2}\right) \frac{d(2 t)}{d t}-2 t \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}}$                                                             [Using quotient rule]

$=\frac{1+t^{2}}{\sqrt{\left(1+t^{2}\right)^{2}-4 t^{2}}} \times \frac{\left(1+t^{2}\right)(2)-(2 t)(2 t)}{\left(1+t^{2}\right)^{2}} \\$

\begin{aligned} & &=\frac{2+2 t^{2}-4 t^{2}}{\sqrt{1+t^{4}+2 t^{2}-4 t^{2}} \times\left(1+t^{2}\right)} \end{aligned}

$=\frac{2-2 t^{2}}{\sqrt{1+t^{4}-2 t^{2}} \times\left(1+t^{2}\right)} \\$

\begin{aligned} & &=\frac{2\left(1-t^{2}\right)}{\sqrt{\left(1-t^{2}\right)^{2}}\left(1+t^{2}\right)} \end{aligned}

$\frac{d x}{d t}=\frac{2}{1+t^{2}} \\$

\begin{aligned} & &y=\tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right) \end{aligned}                                                                                                                              (1)

$\frac{d y}{d t}=\frac{d \tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)}{d t} \\$

\begin{aligned} & &\frac{d y}{d t}=\frac{d \tan ^{-1}\left(\frac{2 t}{1-t^{2}}\right)}{d\left(\frac{2 t}{1-t^{2}}\right)} \times \frac{d\left(\frac{2 t}{1-t^{2}}\right)}{d t} \end{aligned}

$=\frac{1}{1+\left(\frac{2 t}{1-t}\right)} \times \frac{\left(1-t^{2}\right) \frac{d(2 t)}{d t}-(2 t) \frac{d\left(1-t^{2}\right)}{d t}}{\left(1-t^{2}\right)^{2}}$                                                   [Using chain rule]

$=\frac{1}{1+\frac{4 t^{2}}{\left(1-t^{2}\right)^{2}}} \times \frac{\left(1-t^{2}\right) \times 2-(2 t)(-2 t)}{\left(1-t^{2}\right)^{2}} \\$

\begin{aligned} & &=\frac{\left(1-t^{2}\right)^{2}}{\left(1-t^{2}\right)^{2}+4 t^{2}} \times \frac{2-2 t^{2}+4 t^{2}}{\left(1-t^{2}\right)^{2}} \end{aligned}

$=\frac{2+2 t^{2}}{1+t^{4}-2 t^{2}+4 t^{2}} \\$

\begin{aligned} & &=\frac{2(1+(-2))}{1+t^{4}+2 t^{2}} \end{aligned}

$=\frac{2\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}} \\$

\begin{aligned} & &\frac{d y}{d t}=\frac{2}{\left(1+t^{2}\right)} \end{aligned}                                                                                                                            (2)

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

So putting the value of  $\frac{d x}{d t} \text { and } \frac{d y}{d t}$  from the equation (1) and (2) respectively

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

$=\frac{2\left(1+t^{2}\right)}{2\left(1+t^{2}\right)} \\$

\begin{aligned} &\ &\frac{d y}{d x}=1 \end{aligned}

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