The complete solution set of the equation
`sin^(-1) sqrt((1+x)/(2))-sqrt(2-x)=cot^(-1)(tan sqrt(2-x))-sin^(-1) sqrt((1-x)/(2))` is :

Options

A. `[2-(pi^(2))/(4),1]`
B. `[1-(pi^(2))/(4), 1]`
C. `[2-(pi^(2))/(4), 0]`
D. `[-1, 1]`

Answer

To solve the equation

\[ \sin^{-1} \left( \sqrt{\frac{1+x}{2}} \right) - \sqrt{2-x} = \cot^{-1} \left( \tan \sqrt{2-x} \right) - \sin^{-1} \left( \sqrt{\frac{1-x}{2}} \right), \]

we will follow these steps:

### Step 1: Rearranging the Equation
We start by moving the term \(\sin^{-1} \left( \sqrt{\frac{1-x}{2}} \right)\) to the left-hand side:

\[ \sin^{-1} \left( \sqrt{\frac{1+x}{2}} \right) + \sin^{-1} \left( \sqrt{\frac{1-x}{2}} \right) - \sqrt{2-x} = \cot^{-1} \left( \tan \sqrt{2-x} \right). \]

### Step 2: Using the Property of Inverse Sine
Using the property of inverse sine, we can combine the two sine inverse terms:

\[ \sin^{-1} \left( \sqrt{\frac{1+x}{2}} \cdot \sqrt{1 - \frac{1-x}{2}} + \sqrt{\frac{1-x}{2}} \cdot \sqrt{1 - \frac{1+x}{2}} \right) - \sqrt{2-x} = \cot^{-1} \left( \tan \sqrt{2-x} \right). \]

### Step 3: Simplifying the Left Side
The left-hand side simplifies to:

\[ \sin^{-1}(1) - \sqrt{2-x} = \cot^{-1} \left( \tan \sqrt{2-x} \right). \]

Since \(\sin^{-1}(1) = \frac{\pi}{2}\), we have:

\[ \frac{\pi}{2} - \sqrt{2-x} = \cot^{-1} \left( \tan \sqrt{2-x} \right). \]

### Step 4: Understanding Cotangent Inverse
The expression \(\cot^{-1}(\tan \theta)\) simplifies to \(\frac{\pi}{2} - \theta\) when \(\theta\) is in the range \((0, \frac{\pi}{2})\). Thus, we can write:

\[ \frac{\pi}{2} - \sqrt{2-x} = \frac{\pi}{2} - \sqrt{2-x}. \]

### Step 5: Setting Up the Inequalities
For the equation to hold, we need:

\[ \sqrt{2-x} \in (0, \frac{\pi}{2}). \]

This leads to the inequalities:

1. \(2 - x > 0 \Rightarrow x < 2\)
2. \(2 - x < \frac{\pi^2}{4} \Rightarrow x > 2 - \frac{\pi^2}{4}\)

### Step 6: Final Solution Set
Thus, the complete solution set for \(x\) is:

\[ 2 - \frac{\pi^2}{4} < x < 2. \]