If `2tan^(-1)(costheta)=tan^(-1)(2 cosec theta), " then " theta=`

Options

A. `npi-pi/4`
B. `npi+pi/4`
C. `npi+pi/3`
D. `npi - pi/3`

Answer

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

### Step 1: Use the double angle formula for arctangent
We know that:
\[ \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
Thus, we can express \(2\tan^{-1}(\cos \theta)\) as:
\[ \tan^{-1}\left(\frac{2\cos \theta}{1 - \cos^2 \theta}\right) \]
Since \(1 - \cos^2 \theta = \sin^2 \theta\), we can rewrite it as:
\[ \tan^{-1}\left(\frac{2\cos \theta}{\sin^2 \theta}\right) \]

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

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

### Step 4: Substitute \(\csc \theta\)
Recall that \(\csc \theta = \frac{1}{\sin \theta}\), so we can rewrite the right side:
\[ \frac{2\cos \theta}{\sin^2 \theta} = \frac{2}{\sin \theta} \]

### Step 5: Cross-multiply
Cross-multiplying gives us:
\[ 2\cos \theta \cdot \sin \theta = 2 \sin^2 \theta \]

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

### Step 7: Rearrange the equation
Rearranging gives:
\[ \cos \theta \cdot \sin \theta - \sin^2 \theta = 0 \]
Factoring out \(\sin \theta\):
\[ \sin \theta (\cos \theta - \sin \theta) = 0 \]

### Step 8: Solve for \(\theta\)
Setting each factor to zero gives us two cases:
1. \(\sin \theta = 0\) which implies \(\theta = n\pi\) for \(n \in \mathbb{Z}\).
2. \(\cos \theta - \sin \theta = 0\) which implies \(\cos \theta = \sin \theta\) or \(\theta = \frac{\pi}{4} + n\pi\) for \(n \in \mathbb{Z}\).

### Final Answer
Thus, the solutions for \(\theta\) are:
\[ \theta = n\pi \quad \text{or} \quad \theta = \frac{\pi}{4} + n\pi \]