$x = \cos^{- 1}{\left( \frac{1}{\sqrt{1 + t^{2}}} \right),}\ y = \sin^{- 1}{(\frac{t}{\sqrt{1 + t^{2}}}) \Rightarrow \frac{dy}{dx}}\$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
Given, $x = \cos^{- 1}\left( \frac{1}{\sqrt{1 + t^{2}}} ight)$
and
$y = \sin^{- 1}\left( \frac{t}{\sqrt{1 + t^{2}}} ight)$
$\Rightarrow = 	an^{- 1}{t,}$
and $y = 	an^{- 1}t$
$	herefore\ \ y = x \Rightarrow \frac{dy}{dx} = 1$

Q. $x = \cos^{- 1}{\left( \frac{1}{\sqrt{1 + t^{2}}} \right),}\ y = \sin^{- 1}{(\frac{t}{\sqrt{1 + t^{2}}}) \Rightarrow \frac{dy}{dx}}\$ is equal to

Explanation

Reviewed by Dr. Abubacker

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