Prove that `1/2 cos^(-1) ((1-x)/(1+x))=tan^(-1) sqrtx`.

Answer

To prove that

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

we will follow these steps:

### Step 1: Start with the Left-Hand Side (LHS)

We have:

\[ LHS = \frac{1}{2} \cos^{-1} \left( \frac{1-x}{1+x} \right) \]

### Step 2: Use the Double Angle Formula for Tangent

Recall the formula for the double angle of the tangent function:

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

We want to express \( \frac{1}{2} \cos^{-1} \left( \frac{1-x}{1+x} \right) \) in a form that relates to \( \tan^{-1} \).

### Step 3: Set \( x = \sqrt{x} \)

Let \( x = \sqrt{x} \). Then, we can apply the double angle formula:

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

### Step 4: Substitute into the LHS

Substituting back, we have:

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

### Step 5: Conclusion

Thus, we have shown that:

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

This completes the proof.