Question:
If \(y = \cot^{-1}\left[\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} – \sqrt{1-\sin x}}\right]\), where \(0 < x < \frac{\pi}{2}\), then \(\frac{dy}{dx}\) is equal to:
यदि \(y = \cot^{-1}\left[\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} – \sqrt{1-\sin x}}\right]\), जहाँ \(0 < x < \frac{\pi}{2}\), तो \(\frac{dy}{dx}\) का मान किसके बराबर है?
Options:
- 1/2
- 2
- \(\sin x + \cos x\)
- \(\sin x – \cos x\)
- 1/2
- 2
- \(\sin x + \cos x\)
- \(\sin x – \cos x\)
Correct Answer: a)
Explanation:
For \(x \in (0, \pi/2)\), \(x/2 \in (0, \pi/4)\), so \(\cos(x/2) > \sin(x/2) > 0\).
\[\sqrt{1+\sin x} = \cos(x/2) + \sin(x/2)\]
\[\sqrt{1-\sin x} = \cos(x/2) – \sin(x/2)\]
Numerator \(= 2\cos(x/2)\).
Denominator \(= 2\sin(x/2)\).
\[\frac{\text{Numerator}}{\text{Denominator}} = \frac{2\cos(x/2)}{2\sin(x/2)} = \cot\left(\frac{x}{2}\right)\]
Therefore:
\[y = \cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}\]
Differentiating with respect to \(x\):
\[\frac{dy}{dx} = \frac{1}{2}\]
Hence, the correct option is (a).
\(0 < x < \pi/2\) के लिए:
\(\sqrt{1+\sin x} = \cos(x/2) + \sin(x/2)\)
\(\sqrt{1-\sin x} = \cos(x/2) – \sin(x/2)\)
अंश \(= 2\cos(x/2)\) तथा हर \(= 2\sin(x/2)\)।
\[y = \cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}\]
अवकलन करने पर:
\[\frac{dy}{dx} = \frac{1}{2}\]
अतः सही विकल्प (a) है।