Prove that : `(1)/(2) cos^(-1) ((1+ 2 cos x)/( 2+cosx) ) = tan^(-1) ((1)/(sqrt(3)) "tan" (x)/(2))`

Answer

To prove the equation

\[ \frac{1}{2} \cos^{-1} \left( \frac{1 + 2 \cos x}{2 + \cos x} \right) = \tan^{-1} \left( \frac{1}{\sqrt{3}} \tan \left( \frac{x}{2} \right) \right), \]

we will start with the left-hand side (LHS) and manipulate it to reach the right-hand side (RHS).

### Step 1: Rewrite the LHS
We start with the left-hand side:

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

### Step 2: Use the identity for \(\cos^{-1}\)
We know that:

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

Let \( x = \frac{1 + 2 \cos x}{2 + \cos x} \). Thus, we can write:

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

### Step 3: Simplify the expression
Now, we need to simplify \( \sqrt{1 - \left( \frac{1 + 2 \cos x}{2 + \cos x} \right)^2} \) and \( 1 + \frac{1 + 2 \cos x}{2 + \cos x} \).

Calculating \( \left( \frac{1 + 2 \cos x}{2 + \cos x} \right)^2 \):

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

### Step 4: Calculate \( 1 - x^2 \)
Now we calculate \( 1 - x^2 \):

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

Expanding both squares:

\[ (2 + \cos x)^2 = 4 + 4 \cos x + \cos^2 x, \]
\[ (1 + 2 \cos x)^2 = 1 + 4 \cos x + 4 \cos^2 x. \]

Subtracting these:

\[ (2 + \cos x)^2 - (1 + 2 \cos x)^2 = (4 + 4 \cos x + \cos^2 x) - (1 + 4 \cos x + 4 \cos^2 x) = 3 - 3 \cos^2 x = 3(1 - \cos^2 x) = 3 \sin^2 x. \]

Thus,

\[ \sqrt{1 - x^2} = \sqrt{\frac{3 \sin^2 x}{(2 + \cos x)^2}} = \frac{\sqrt{3} \sin x}{2 + \cos x}. \]

### Step 5: Substitute back into the expression
Now substituting back into our expression for \(\tan^{-1}\):

\[ \tan^{-1} \left( \frac{\sqrt{3} \sin x}{2 + \cos x} \cdot \frac{1}{1 + \frac{1 + 2 \cos x}{2 + \cos x}} \right). \]

The denominator simplifies to:

\[ 1 + \frac{1 + 2 \cos x}{2 + \cos x} = \frac{(2 + \cos x) + (1 + 2 \cos x)}{2 + \cos x} = \frac{3 + 3 \cos x}{2 + \cos x} = \frac{3(1 + \cos x)}{2 + \cos x}. \]

### Step 6: Final simplification
Thus, we have:

\[ \text{LHS} = \tan^{-1} \left( \frac{\sqrt{3} \sin x}{2 + \cos x} \cdot \frac{2 + \cos x}{3(1 + \cos x)} \right) = \tan^{-1} \left( \frac{\sqrt{3} \sin x}{3(1 + \cos x)} \right). \]

Using the double angle identity for sine, \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \) and \( 1 + \cos x = 2 \cos^2 \frac{x}{2} \):

\[ \text{LHS} = \tan^{-1} \left( \frac{\sqrt{3} \cdot 2 \sin \frac{x}{2} \cos \frac{x}{2}}{3 \cdot 2 \cos^2 \frac{x}{2}} \right) = \tan^{-1} \left( \frac{\sqrt{3}}{3} \tan \frac{x}{2} \right). \]

Since \( \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \), we have:

\[ \text{LHS} = \tan^{-1} \left( \frac{1}{\sqrt{3}} \tan \frac{x}{2} \right). \]

### Conclusion
Thus, we have shown that:

\[ \frac{1}{2} \cos^{-1} \left( \frac{1 + 2 \cos x}{2 + \cos x} \right) = \tan^{-1} \left( \frac{1}{\sqrt{3}} \tan \left( \frac{x}{2} \right) \right), \]

which proves the statement.