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: Test a value
Take $x = \frac\pi3$: $\sin x = 0.866$, $\sqrt{1-\sin x} = 0.366$, $\sqrt{1+\sin x} = 1.366$. The fraction is $\frac{1.732}{-1} = -1.732$.

Step 2: Evaluate
$\cot^{-1}(-1.732)$ lies in $(0,\pi)$ and equals $\frac{5\pi}{6}$.

Step 3: Compare
Option (D): $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$. It matches. Option (C): $\frac\pi2-\frac\pi6 = \frac\pi3$ does not.

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