Question

Evaluate: \[ \cot^{-1}\!\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) \]

• A. \(\frac{x}{3}\)
• B. \(\frac{x}{4}\)
• C. \(1\)
• D. \(\frac{x}{2}\)

Hint:

Expressions with \(\sqrt{1 \pm \sin x}\) often simplify using half-angle identities: \(1 \pm \sin x = (\sin\frac{x}{2} \pm \cos\frac{x}{2})^2\).

Answer 1

The Correct Option is D

Concept: Use half-angle identities to simplify the expression inside the inverse cotangent.

Step 1: Express \(\sqrt{1 \pm \sin x}\) in terms of half-angles. \[ 1 + \sin x = \left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2 \] \[ 1 - \sin x = \left(\sin\frac{x}{2} - \cos\frac{x}{2}\right)^2 \] For \(x \in (0, \pi/2)\) (appropriate domain), \(\sin\frac{x}{2} + \cos\frac{x}{2}>0\) and \(\cos\frac{x}{2}>\sin\frac{x}{2}\). So: \[ \sqrt{1+\sin x} = \sin\frac{x}{2} + \cos\frac{x}{2} \] \[ \sqrt{1-\sin x} = \cos\frac{x}{2} - \sin\frac{x}{2} \]

Step 2: Simplify the fraction. Numerator: \[ \sqrt{1+\sin x} + \sqrt{1-\sin x} = (\sin\frac{x}{2} + \cos\frac{x}{2}) + (\cos\frac{x}{2} - \sin\frac{x}{2}) = 2\cos\frac{x}{2} \] Denominator: \[ \sqrt{1+\sin x} - \sqrt{1-\sin x} = (\sin\frac{x}{2} + \cos\frac{x}{2}) - (\cos\frac{x}{2} - \sin\frac{x}{2}) = 2\sin\frac{x}{2} \] Thus: \[ \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} = \frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot\frac{x}{2} \]

Step 3: Apply inverse cotangent. \[ \cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2} \] (provided \(\frac{x}{2}\) is in the principal range of \(\cot^{-1}\), which is \((0, \pi)\)).