Question

If $y = \operatorname{cosec}^{-1} \left[ \frac{\sqrt{x}+1}{\sqrt{x}-1} \right] + \cos^{-1} \left[ \frac{\sqrt{x}-1}{\sqrt{x}+1} \right]$, then $\frac{dy}{dx} =$

• A. 0
• B. 1
• C. $\frac{2}{x+1}$
• D. $\frac{1}{2(x-1)}$

Hint:

Whenever you see inverse trigonometric functions added together, always check if the arguments can be manipulated to match. If they match, the sum is almost always $\frac{\pi}{2}$!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the derivative of a complex trigonometric function.

Step 2: Key Formula or Approach:
Use the trigonometric identity $\operatorname{cosec}^{-1}(\theta) = \sin^{-1}(1/\theta)$.
The sum of inverse functions identity: $\sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2}$.

Step 3: Detailed Explanation:
$y = \operatorname{cosec}^{-1} \left[ \frac{\sqrt{x}+1}{\sqrt{x}-1} \right] + \cos^{-1} \left[ \frac{\sqrt{x}-1}{\sqrt{x}+1} \right]$
Convert $\operatorname{cosec}^{-1}$ to $\sin^{-1}$:
$y = \sin^{-1} \left[ \frac{\sqrt{x}-1}{\sqrt{x}+1} \right] + \cos^{-1} \left[ \frac{\sqrt{x}-1}{\sqrt{x}+1} \right]$
Since the arguments are identical, let $u = \frac{\sqrt{x}-1}{\sqrt{x}+1}$.
$y = \sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2}$.
The derivative of any constant (like $\frac{\pi}{2}$) is zero.
$\frac{dy}{dx} = 0$.

Step 4: Final Answer:
The derivative is 0, matching option (A).