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

Options

A. `1/2`
B. `(1)/sqrt(3)`
C. `sqrt(3)`
D. 2

Explanation

Let's solve the given equation step by step:

Given:
\[ \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2} \tan^{-1}(x) \]

Step 1: Use the formula for the difference of two inverse tangents:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \]

In our case, we can rewrite \(\tan^{-1}\left(\frac{1-x}{1+x}\right)\) using the above formula:
\[ \tan^{-1}(1) - \tan^{-1}(x) = \tan^{-1}\left(\frac{1-x}{1+x}\right) \]

Therefore:
\[ \tan^{-1}(1) - \tan^{-1}(x) = \frac{1}{2} \tan^{-1}(x) \]

Step 2: Let \( \theta = \tan^{-1}(x) \). Then the equation becomes:
\[ \tan^{-1}(1) - \theta = \frac{1}{2} \theta \]

Step 3: We know that \( \tan^{-1}(1) = \frac{\pi}{4} \). Substitute this into the equation:
\[ \frac{\pi}{4} - \theta = \frac{1}{2} \theta \]

Step 4: Solve for \( \theta \):
\[ \frac{\pi}{4} = \frac{3}{2} \theta \]
\[ \theta = \frac{\pi}{4} \cdot \frac{2}{3} \]
\[ \theta = \frac{\pi}{6} \]

Step 5: Recall that \( \theta = \tan^{-1}(x) \). Therefore:
\[ \tan^{-1}(x) = \frac{\pi}{6} \]

Step 6: Find \( x \):
\[ x = \tan\left(\frac{\pi}{6}\right) \]
\[ x = \frac{1}{\sqrt{3}} \]

Thus, the value of \( x \) is:
\[ x = \frac{1}{\sqrt{3}} \]