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: Read the function.
We have $y=\operatorname{cosec}^{-1}\!\left[\frac{\sqrt{x}+1}{\sqrt{x}-1}\right]+\cos^{-1}\!\left[\frac{\sqrt{x}-1}{\sqrt{x}+1}\right]$ and we want $\frac{dy}{dx}$.
Step 2: Convert the cosec term.
A handy identity is $\operatorname{cosec}^{-1}(\theta)=\sin^{-1}(1/\theta)$. So the first term becomes $\sin^{-1}\!\left[\frac{\sqrt{x}-1}{\sqrt{x}+1}\right]$ after flipping the fraction.
Step 3: Spot the matching arguments.
Now both inverse functions hold the same inside value. Let $u=\frac{\sqrt{x}-1}{\sqrt{x}+1}$.
Step 4: Use the sum identity.
For any allowed value, $\sin^{-1}(u)+\cos^{-1}(u)=\frac{\pi}{2}$. So $y=\frac{\pi}{2}$.
Step 5: See that y is constant.
The whole expression collapses to a fixed number, $\frac{\pi}{2}$. It does not depend on $x$ at all.
Step 6: Differentiate.
The derivative of a constant is zero, so $\frac{dy}{dx}=0$. This is option (A).
\[ \boxed{\,\dfrac{dy}{dx}=0\,} \]