Question

If $x = \tan^{-1} \left\{ \frac{\sqrt{1+t^2}-1}{t} \right\}, y = \cos^{-1} \left\{ \frac{1-t^2}{1+t^2} \right\}$, then $\frac{dy}{dx}$ is equal to

• A. 2
• B. $\frac{1}{2}$
• C. 4
• D. $\frac{1}{4}$

Hint:

Always look for trigonometric substitutions in inverse trig functions. For $1+t^2$, use $t = \tan \theta$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This is a parametric differentiation problem. We can simplify the expressions for \( x \) and \( y \) using trigonometric substitutions.
Step 3: Detailed Explanation:
Let \( t = \tan \theta \). Then \( \theta = \tan^{-1} t \).
For \( x \):
\[ x = \tan^{-1} \left( \frac{\sec \theta - 1}{\tan \theta} \right) = \tan^{-1} \left( \frac{1 - \cos \theta}{\sin \theta} \right) = \tan^{-1} \left( \tan \frac{\theta}{2} \right) = \frac{\theta}{2} \]
\[ x = \frac{1}{2} \tan^{-1} t \]
For \( y \):
\[ y = \cos^{-1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) = \cos^{-1} (\cos 2\theta) = 2\theta \]
\[ y = 2 \tan^{-1} t \]
Relating \( y \) and \( x \):
\[ y = 4 \times \left( \frac{1}{2} \tan^{-1} t \right) = 4x \]
Differentiating \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 4 \]
Step 4: Final Answer:
The derivative \( \frac{dy}{dx} \) is \( 4 \).