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We are given two equations for \(x\) and \(y\) in terms of a parameter \(t\):

\[
x = \csc^{-1}\left(\frac{1 + t^2}{2t}\right)
\]
and
\[
y = \tan^{-1}\left(\frac{3t - t^3}{1 - 3t^2}\right).
\]

We need to find \(\frac{dy}{dx}\). This involves using the chain rule, i.e.,

\[
\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}.
\]

### Step 1: Differentiate \(x\) with respect to \(t\)

We are given:
\[
x = \csc^{-1}\left(\frac{1 + t^2}{2t}\right).
\]

The derivative of \(\csc^{-1}(u)\) with respect to \(t\) is:

\[
\frac{d}{dt} \csc^{-1}(u) = -\frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dt}.
\]

Here, \( u = \frac{1 + t^2}{2t} \), so we first need to find \(\frac{du}{dt}\):

\[
u = \frac{1 + t^2}{2t}.
\]
Using the quotient rule:

\[
\frac{du}{dt} = \frac{(2t)(2t) - (1 + t^2)(2)}{(2t)^2} = \frac{4t^2 - 2 - 2t^2}{4t^2} = \frac{2t^2 - 2}{4t^2} = \frac{1 - 1/t^2}{2}.
\]

Thus,

\[
\frac{dx}{dt} = -\frac{1}{\

Find dy/dx when x = cosec^{-1}((1 + t^2) / (2t)) and y = tan^{-1}((3t - t^3) / (1 - 3t^2)).

Step-by-step solution to finding dy/dx when x = cosec^{-1}((1 + t^2) / (2t)) and y = tan^{-1}((3t - t^3) / (1 - 3t^2)). The solution covers key concepts such as inverse trigonometric functions, chain rule, and quotient rule.

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