Question:
\( x = \cos^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right), \; y = \sin^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right) \implies \frac{dy}{dx} \) is equal to
\( x = \cos^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right), \; y = \sin^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right) \implies \frac{dy}{dx} \) is equal to
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Trigonometric substitution ($t = \tan \theta$) simplifies many parametric derivatives.
Updated On: Apr 10, 2026
- 0
- $\tan t$
- 1
- $\sin t \cos t$
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The Correct Option is C
Solution and Explanation
Step 1: Substitution
Let $t = \tan \theta$. Then $\frac{1}{\sqrt{1+t^2}} = \frac{1}{\sec \theta} = \cos \theta$ and $\frac{t}{\sqrt{1+t^2}} = \frac{\tan \theta}{\sec \theta} = \sin \theta$.
Step 2: Simplify Functions
$x = \cos^{-1}(\cos \theta) = \theta = \tan^{-1} t$. $y = \sin^{-1}(\sin \theta) = \theta = \tan^{-1} t$.
Step 3: Derivative
Since $y = x$, then $\frac{dy}{dx} = 1$.
Final Answer: (c)
Let $t = \tan \theta$. Then $\frac{1}{\sqrt{1+t^2}} = \frac{1}{\sec \theta} = \cos \theta$ and $\frac{t}{\sqrt{1+t^2}} = \frac{\tan \theta}{\sec \theta} = \sin \theta$.
Step 2: Simplify Functions
$x = \cos^{-1}(\cos \theta) = \theta = \tan^{-1} t$. $y = \sin^{-1}(\sin \theta) = \theta = \tan^{-1} t$.
Step 3: Derivative
Since $y = x$, then $\frac{dy}{dx} = 1$.
Final Answer: (c)
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