Question

The value of \(cot^{-1}[\frac{\sqrt{1-sinx}+\sqrt{1+sinx}}{\sqrt{1-sinx}-\sqrt{1+sinx}}]\), where \(x\in (0,\frac{π}{2})\) is...

• A. \(π-x\)
• B. \(2π-x\)
• C. \(\frac{π}{2}-\frac{x}{2}\)
• D. \(π-\frac{x}{2}\)

Hint:

Use half-angle forms: 1 plus or minus sin x equals (cos(x/2) plus or minus sin(x/2)) squared.

Answer 1

The Correct Option is D

Step 1: Rewrite the roots
\(1\pm\sin x = \left(\cos\frac x2 \pm \sin\frac x2\right)^2\). For \(x\in(0,\frac\pi2)\), \(\frac x2 \in (0,\frac\pi4)\), so \(\cos\frac x2 > \sin\frac x2\). Hence \(\sqrt{1+\sin x} = \cos\frac x2+\sin\frac x2\) and \(\sqrt{1-\sin x} = \cos\frac x2-\sin\frac x2\).

Step 2: Evaluate the fraction
\[ \frac{(c-s)+(c+s)}{(c-s)-(c+s)} = \frac{2c}{-2s} = -\cot\frac x2 \]

Step 3: Apply the inverse
\(\cot^{-1}(-y) = \pi - \cot^{-1}y\) because the principal range is \((0,\pi)\). So the result is \(\pi - \frac x2\).

Step 4: Result
Option (D). Options (A) and (B) have the wrong coefficient of \(x\). Option (C) forgets the negative sign.

Final Answer:
The value is pi - x/2. \[ \boxed{\text{(D)}\ \pi-\frac x2} \]