If `tan^(-1) (sqrt(( 1 +x^(2)) -1)/x) = lambda tan^(-1)x` then the value of `lambda` is

Options

A. 2
B. `1/2`
C. 1
D. none of these

Answer

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

### Step 1: Substitute \( x \) with \( \tan(\theta) \)
Let \( x = \tan(\theta) \). Then we can express \( 1 + x^2 \) as:
\[ 1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta) \]

### Step 2: Rewrite the left-hand side
Now, substituting \( x \) in the left-hand side gives:
\[ \tan^{-1} \left( \frac{\sqrt{\sec^2(\theta)} - 1}{\tan(\theta)} \right) \]
Since \( \sqrt{\sec^2(\theta)} = \sec(\theta) \), we have:
\[ \tan^{-1} \left( \frac{\sec(\theta) - 1}{\tan(\theta)} \right) \]

### Step 3: Simplify the expression
Using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we can rewrite:
\[ \frac{\sec(\theta) - 1}{\tan(\theta)} = \frac{\frac{1}{\cos(\theta)} - 1}{\frac{\sin(\theta)}{\cos(\theta)}} = \frac{1 - \cos(\theta)}{\sin(\theta)} \]

### Step 4: Use the identity for \( 1 - \cos(\theta) \)
We can use the identity \( 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \):
\[ \frac{1 - \cos(\theta)}{\sin(\theta)} = \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} = \tan\left(\frac{\theta}{2}\right) \]

### Step 5: Rewrite the left-hand side
Thus, we have:
\[ \tan^{-1} \left( \tan\left(\frac{\theta}{2}\right) \right) = \frac{\theta}{2} \]

### Step 6: Relate \( \theta \) back to \( x \)
Since \( \theta = \tan^{-1}(x) \), we can substitute back:
\[ \frac{\tan^{-1}(x)}{2} = \lambda \tan^{-1}(x) \]

### Step 7: Solve for \( \lambda \)
Dividing both sides by \( \tan^{-1}(x) \) (assuming \( \tan^{-1}(x) \neq 0 \)):
\[ \frac{1}{2} = \lambda \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \lambda = \frac{1}{2} \]

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