Solve the equation `2 tan^(-1) (cos x) = tan^(-1) (2 cosec x)`

Explanation

To solve the equation \( 2 \tan^{-1}(\cos x) = \tan^{-1}(2 \csc x) \), we can follow these steps:

### Step 1: Use the double angle formula for inverse tangent
We know that:
\[ 2 \tan^{-1}(y) = \tan^{-1}\left(\frac{2y}{1 - y^2}\right) \]
Applying this to our equation:
\[ 2 \tan^{-1}(\cos x) = \tan^{-1}\left(\frac{2 \cos x}{1 - \cos^2 x}\right) \]
Since \(1 - \cos^2 x = \sin^2 x\), we can rewrite this as:
\[ 2 \tan^{-1}(\cos x) = \tan^{-1}\left(\frac{2 \cos x}{\sin^2 x}\right) \]

### Step 2: Set the two sides equal
Now we have:
\[ \tan^{-1}\left(\frac{2 \cos x}{\sin^2 x}\right) = \tan^{-1}(2 \csc x) \]

### Step 3: Remove the inverse tangent
Since the tangent function is one-to-one, we can equate the arguments:
\[ \frac{2 \cos x}{\sin^2 x} = 2 \csc x \]

### Step 4: Rewrite cosecant in terms of sine
Recall that \(\csc x = \frac{1}{\sin x}\), so we can rewrite the equation:
\[ \frac{2 \cos x}{\sin^2 x} = \frac{2}{\sin x} \]

### Step 5: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 2 \cos x \cdot \sin x = 2 \sin^2 x \]

### Step 6: Simplify the equation
Dividing both sides by 2 (assuming \(2 \neq 0\)):
\[ \cos x \cdot \sin x = \sin^2 x \]

### Step 7: Rearranging the equation
We can rearrange this to:
\[ \cos x \cdot \sin x - \sin^2 x = 0 \]
Factoring out \(\sin x\):
\[ \sin x (\cos x - \sin x) = 0 \]

### Step 8: Solve for \(x\)
This gives us two cases:
1. \(\sin x = 0\)
2. \(\cos x - \sin x = 0\)

**Case 1:** \(\sin x = 0\)
- This occurs at \(x = n\pi\), where \(n\) is any integer.

**Case 2:** \(\cos x = \sin x\)
- This occurs at \(x = \frac{\pi}{4} + n\pi\), where \(n\) is any integer.

### Final Solution
Thus, the solutions to the equation \(2 \tan^{-1}(\cos x) = \tan^{-1}(2 \csc x)\) are:
\[ x = n\pi \quad \text{and} \quad x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]