`tan[(sqrt(1+x^(2))-1)/x]`=

Options

A. `tan^(-1)x`
B. `1/2tan^(-1)x`
C. `2.tan^(-1)x`
D. none of these

Answer

To solve the expression \( \tan\left(\frac{\sqrt{1+x^2}-1}{x}\right) \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \tan\left(\frac{\sqrt{1+x^2}-1}{x}\right) \]

### Step 2: Substitute \( x = \tan(\theta) \)
Let \( x = \tan(\theta) \). Then, we have:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sec(\theta) \]
So, the expression becomes:
\[ \tan\left(\frac{\sec(\theta) - 1}{\tan(\theta)}\right) \]

### Step 3: Simplify the expression
Now, we simplify \( \frac{\sec(\theta) - 1}{\tan(\theta)} \):
\[ \frac{\sec(\theta) - 1}{\tan(\theta)} = \frac{\sec(\theta) - 1}{\frac{\sin(\theta)}{\cos(\theta)}} = \frac{(\sec(\theta) - 1) \cos(\theta)}{\sin(\theta)} \]
Since \( \sec(\theta) = \frac{1}{\cos(\theta)} \), we can rewrite this as:
\[ \frac{1 - \cos(\theta)}{\sin(\theta)} \]

### Step 4: Use trigonometric identities
Using the identity \( 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \), we have:
\[ \frac{1 - \cos(\theta)}{\sin(\theta)} = \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} = \tan\left(\frac{\theta}{2}\right) \]

### Step 5: Substitute back
Now substituting back, we get:
\[ \tan\left(\frac{\sec(\theta) - 1}{\tan(\theta)}\right) = \tan\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2} \]

### Step 6: Substitute \( \theta \) back in terms of \( x \)
Since we have \( \theta = \tan^{-1}(x) \), we can write:
\[ \frac{\theta}{2} = \frac{1}{2} \tan^{-1}(x) \]

### Final Answer
Thus, the final result is:
\[ \tan\left(\frac{\sqrt{1+x^2}-1}{x}\right) = \frac{1}{2} \tan^{-1}(x) \]