Is `tan^(-1) (sqrt((1-x^(2))/(1+x^(2))) ) = 1/2 cos^(-1) x ` true ?

Answer

To determine whether the equation \( \tan^{-1} \left( \sqrt{\frac{1-x^2}{1+x^2}} \right) = \frac{1}{2} \cos^{-1} x \) is true, we will analyze both sides step by step.

### Step 1: Understanding the left side
We start with the left side of the equation:
\[ y = \tan^{-1} \left( \sqrt{\frac{1-x^2}{1+x^2}} \right) \]
This expression involves the inverse tangent function.

### Step 2: Simplifying the expression
We can use the identity for tangent:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Here, we can express \( \sqrt{\frac{1-x^2}{1+x^2}} \) in terms of sine and cosine. We know that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Let \( x = \cos \theta \). Then:
\[ 1 - x^2 = \sin^2 \theta \quad \text{and} \quad 1 + x^2 = 1 + \cos^2 \theta = 1 + \cos^2 \theta \]
Thus, we can rewrite:
\[ \sqrt{\frac{1-x^2}{1+x^2}} = \sqrt{\frac{\sin^2 \theta}{1 + \cos^2 \theta}} = \frac{\sin \theta}{\sqrt{1 + \cos^2 \theta}} \]

### Step 3: Using the half-angle identity
Using the half-angle identities, we know:
\[ \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} \]
We can relate this to our expression:
\[ \tan^{-1} \left( \frac{\sin \theta}{\sqrt{1 + \cos^2 \theta}} \right) \]

### Step 4: Analyzing the right side
Now we analyze the right side:
\[ \frac{1}{2} \cos^{-1} x \]
Using the substitution \( x = \cos \theta \), we have:
\[ \cos^{-1} x = \theta \implies \frac{1}{2} \cos^{-1} x = \frac{\theta}{2} \]

### Step 5: Equating both sides
Now we equate both sides:
\[ \tan^{-1} \left( \sqrt{\frac{1-x^2}{1+x^2}} \right) = \frac{\theta}{2} \]
From the previous steps, we can see that:
\[ \tan^{-1} \left( \tan \frac{\theta}{2} \right) = \frac{\theta}{2} \]
This indicates that both sides are equal under the condition that \( \theta \) is in the appropriate range.

### Conclusion
Thus, we conclude that:
\[ \tan^{-1} \left( \sqrt{\frac{1-x^2}{1+x^2}} \right) = \frac{1}{2} \cos^{-1} x \]
is indeed **true**.