Q68 / 150 · MHT CET 2026 (16 April Shift 2) · Mathematics
The value of $\cot^{-1}\left[\dfrac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}}\right]$, where $x \in \left(0, \dfrac{\pi}{2}\right)$ is...
- A$\pi - x$
- B$2\pi - x$
- C$\dfrac{\pi}{2} - \dfrac{x}{2}$
- Correct answerD$\pi - \dfrac{x}{2}$
Solution
We know that $1 \pm \sin x = \left(\cos \dfrac{x}{2} \pm \sin \dfrac{x}{2}\right)^2$.
Since $x \in \left(0, \dfrac{\pi}{2}\right)$, we have $\dfrac{x}{2} \in \left(0, \dfrac{\pi}{4}\right)$.
In this interval, $\cos \dfrac{x}{2} > \sin \dfrac{x}{2} > 0$.
Therefore, $\sqrt{1 - \sin x} = \cos \dfrac{x}{2} - \sin \dfrac{x}{2}$ and $\sqrt{1 + \sin x} = \cos \dfrac{x}{2} + \sin \dfrac{x}{2}$.
Substituting these values into the given expression:
$\dfrac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} = \dfrac{\left(\cos \dfrac{x}{2} - \sin \dfrac{x}{2}\right) + \left(\cos \dfrac{x}{2} + \sin \dfrac{x}{2}\right)}{\left(\cos \dfrac{x}{2} - \sin \dfrac{x}{2}\right) - \left(\cos \dfrac{x}{2} + \sin \dfrac{x}{2}\right)}$
$= \dfrac{2\cos \dfrac{x}{2}}{-2\sin \dfrac{x}{2}} = -\cot \dfrac{x}{2}$
The given expression becomes $\cot^{-1}\left(-\cot \dfrac{x}{2}\right)$.
Using the property $\cot^{-1}(-y) = \pi - \cot^{-1}y$:
$\cot^{-1}\left(-\cot \dfrac{x}{2}\right) = \pi - \cot^{-1}\left(\cot \dfrac{x}{2}\right)$
Since $\dfrac{x}{2} \in \left(0, \dfrac{\pi}{4}\right)$, $\cot^{-1}\left(\cot \dfrac{x}{2}\right) = \dfrac{x}{2}$.
Thus, the value is $\pi - \dfrac{x}{2}$.
Answer: $\pi - \dfrac{x}{2}$
Since $x \in \left(0, \dfrac{\pi}{2}\right)$, we have $\dfrac{x}{2} \in \left(0, \dfrac{\pi}{4}\right)$.
In this interval, $\cos \dfrac{x}{2} > \sin \dfrac{x}{2} > 0$.
Therefore, $\sqrt{1 - \sin x} = \cos \dfrac{x}{2} - \sin \dfrac{x}{2}$ and $\sqrt{1 + \sin x} = \cos \dfrac{x}{2} + \sin \dfrac{x}{2}$.
Substituting these values into the given expression:
$\dfrac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} = \dfrac{\left(\cos \dfrac{x}{2} - \sin \dfrac{x}{2}\right) + \left(\cos \dfrac{x}{2} + \sin \dfrac{x}{2}\right)}{\left(\cos \dfrac{x}{2} - \sin \dfrac{x}{2}\right) - \left(\cos \dfrac{x}{2} + \sin \dfrac{x}{2}\right)}$
$= \dfrac{2\cos \dfrac{x}{2}}{-2\sin \dfrac{x}{2}} = -\cot \dfrac{x}{2}$
The given expression becomes $\cot^{-1}\left(-\cot \dfrac{x}{2}\right)$.
Using the property $\cot^{-1}(-y) = \pi - \cot^{-1}y$:
$\cot^{-1}\left(-\cot \dfrac{x}{2}\right) = \pi - \cot^{-1}\left(\cot \dfrac{x}{2}\right)$
Since $\dfrac{x}{2} \in \left(0, \dfrac{\pi}{4}\right)$, $\cot^{-1}\left(\cot \dfrac{x}{2}\right) = \dfrac{x}{2}$.
Thus, the value is $\pi - \dfrac{x}{2}$.
Answer: $\pi - \dfrac{x}{2}$
Hence, the correct answer is option: 'D'.