If `sin^(-1) sqrt(x^2+2x+1sec^(-1)) sqrt(x^2+2x+1) = pi/2, x ne 0` then the value of `2sec^(-1) x/2+sin^(-1) x/2=`

Options

A. `-pi/2`
B. `-(3pi)/2, pi/2`
C. `(3pi)/2`
D. `-(3pi)/2`

Answer

To solve the given problem step by step, we start with the equation:

\[ \sin^{-1}(\sqrt{x^2 + 2x + 1}) + \sec^{-1}(\sqrt{x^2 + 2x + 1}) = \frac{\pi}{2} \]

### Step 1: Simplify the expression inside the inverse functions

Notice that:

\[ \sqrt{x^2 + 2x + 1} = \sqrt{(x + 1)^2} = |x + 1| \]

Since \(x \neq 0\), we consider two cases for \(x + 1\): when \(x + 1 \geq 0\) and \(x + 1 < 0\).

### Step 2: Analyze the cases

1. **Case 1**: \(x + 1 \geq 0\) (i.e., \(x \geq -1\))
- Here, \(|x + 1| = x + 1\).
- The equation becomes:
\[ \sin^{-1}(x + 1) + \sec^{-1}(x + 1) = \frac{\pi}{2} \]

2. **Case 2**: \(x + 1 < 0\) (i.e., \(x < -1\))
- Here, \(|x + 1| = -(x + 1) = -x - 1\).
- The equation becomes:
\[ \sin^{-1}(-x - 1) + \sec^{-1}(-x - 1) = \frac{\pi}{2} \]

### Step 3: Use the identity

From the identity \(\sin^{-1}(y) + \sec^{-1}(y) = \frac{\pi}{2}\), we can conclude that:

\[ \sec^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y) \]

This implies that:

\[ y = \sqrt{x^2 + 2x + 1} = |x + 1| \]

### Step 4: Solve for \(x\)

From the equation:

\[ \sqrt{x^2 + 2x + 1} = 1 \]

Squaring both sides gives:

\[ x^2 + 2x + 1 = 1 \]

This simplifies to:

\[ x^2 + 2x = 0 \]

Factoring out \(x\):

\[ x(x + 2) = 0 \]

Thus, \(x = 0\) or \(x = -2\). Since \(x \neq 0\), we have:

\[ x = -2 \]

### Step 5: Find the required expression

Now we need to evaluate:

\[ 2 \sec^{-1}\left(\frac{x}{2}\right) + \sin^{-1}\left(\frac{x}{2}\right) \]

Substituting \(x = -2\):

\[ 2 \sec^{-1}\left(\frac{-2}{2}\right) + \sin^{-1}\left(\frac{-2}{2}\right) = 2 \sec^{-1}(-1) + \sin^{-1}(-1) \]

### Step 6: Calculate the values

- \(\sec^{-1}(-1) = \pi\) (since \(\sec(\pi) = -1\))
- \(\sin^{-1}(-1) = -\frac{\pi}{2}\)

Thus, we have:

\[ 2 \cdot \pi + \left(-\frac{\pi}{2}\right) = 2\pi - \frac{\pi}{2} = \frac{4\pi}{2} - \frac{\pi}{2} = \frac{3\pi}{2} \]

### Final Answer

The value of \(2 \sec^{-1}\left(\frac{x}{2}\right) + \sin^{-1}\left(\frac{x}{2}\right)\) is:

\[ \frac{3\pi}{2} \]

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