If `2tan^(-1)(cosx)=tan^(-1)(2cosecx)`, then x=

Options

A. `(pi)/(4)`
B. `(-pi)/(4)`
C. `(pi)/(3)`
D. `(pi)/(6)`

Answer

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, we set \( y = \cos x \):
\[ 2\tan^{-1}(\cos x) = \tan^{-1}\left(\frac{2\cos x}{1 - \cos^2 x}\right) \]
Since \( 1 - \cos^2 x = \sin^2 x \), we can rewrite it as:
\[ \tan^{-1}\left(\frac{2\cos x}{\sin^2 x}\right) \]

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

### Step 3: Rewrite \( \csc x \)
Recall that \( \csc x = \frac{1}{\sin x} \), so we can rewrite the right side:
\[ \frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x} \]

### Step 4: Cross-multiply
Cross-multiplying gives:
\[ 2\cos x \cdot \sin x = 2\sin^2 x \]

### Step 5: Simplify the equation
We can simplify this to:
\[ \cos x \cdot \sin x = \sin^2 x \]
Dividing both sides by \( \sin x \) (assuming \( \sin x \neq 0 \)):
\[ \cos x = \sin x \]

### Step 6: Solve for \( x \)
The equation \( \cos x = \sin x \) implies:
\[ \tan x = 1 \]
Thus, the general solution for this is:
\[ x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
For the principal value, we have:
\[ x = \frac{\pi}{4} \]

### Conclusion
The solution to the equation \( 2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x) \) is:
\[ \boxed{\frac{\pi}{4}} \]