Find the magnitude of angle A, if :
`tanA-2cosAtanA+2cosA-1=0`

Answer

To solve the equation \( \tan A - 2 \cos A \tan A + 2 \cos A - 1 = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
Start with the given equation:
\[ \tan A - 2 \cos A \tan A + 2 \cos A - 1 = 0 \]
We can rearrange it as:
\[ \tan A (1 - 2 \cos A) + (2 \cos A - 1) = 0 \]

### Step 2: Factor out common terms
Now, we can factor out the common terms:
\[ (1 - 2 \cos A)(\tan A - 1) = 0 \]

### Step 3: Set each factor to zero
This gives us two equations to solve:
1. \( 1 - 2 \cos A = 0 \)
2. \( \tan A - 1 = 0 \)

### Step 4: Solve the first equation
From the first equation:
\[ 1 - 2 \cos A = 0 \implies 2 \cos A = 1 \implies \cos A = \frac{1}{2} \]
The angles for which \( \cos A = \frac{1}{2} \) are:
\[ A = 60^\circ \text{ or } A = 300^\circ \]

### Step 5: Solve the second equation
From the second equation:
\[ \tan A - 1 = 0 \implies \tan A = 1 \]
The angle for which \( \tan A = 1 \) is:
\[ A = 45^\circ \text{ or } A = 225^\circ \]

### Step 6: List all possible angles
From both equations, we have the possible angles:
1. From \( \cos A = \frac{1}{2} \): \( A = 60^\circ, 300^\circ \)
2. From \( \tan A = 1 \): \( A = 45^\circ, 225^\circ \)

### Step 7: Find the magnitude of angle A
Since the question asks for the magnitude of angle A, we consider only the angles in the range \( 0^\circ \) to \( 360^\circ \):
- Possible angles are \( 60^\circ, 300^\circ, 45^\circ, 225^\circ \).

Thus, the magnitudes of angle A are:
\[ A = 60^\circ, 45^\circ, 300^\circ, 225^\circ \]

### Final Answer
The magnitudes of angle A are \( 60^\circ \) and \( 45^\circ \) (as standard angles).

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