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: Simplify $x$
Substitute $t = \tan \theta$.
$x = \tan^{-1} \left( \frac{\sec \theta - 1}{\tan \theta} \right) = \tan^{-1} \left( \frac{1 - \cos \theta}{\sin \theta} \right) = \tan^{-1} (\tan \theta/2) = \theta/2 = \frac{1}{2}\tan^{-1} t$.

Step 2: Simplify $y$
Substitute $t = \tan \theta$.
$y = \cos^{-1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) = \cos^{-1} (\cos 2\theta) = 2\theta = 2\tan^{-1} t$.

Step 3: Calculation
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{2}{1+t^2}}{\frac{1/2}{1+t^2}} = 2 \times 2 = 4$.
Final Answer: (C)