If `2 sec 2 alpha = tan beta + cot beta,` then one of the values of `alpha + beta ` is

Options

A. `pi/4`
B. `pi/2`
C. `pi`
D. `2pi`

Answer

To solve the equation \(2 \sec 2\alpha = \tan \beta + \cot \beta\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2 \sec 2\alpha = \tan \beta + \cot \beta \]
Using the identities \(\sec x = \frac{1}{\cos x}\), \(\tan x = \frac{\sin x}{\cos x}\), and \(\cot x = \frac{\cos x}{\sin x}\), we can rewrite the right-hand side:
\[ \tan \beta + \cot \beta = \frac{\sin \beta}{\cos \beta} + \frac{\cos \beta}{\sin \beta} \]
Combining these fractions gives:
\[ \tan \beta + \cot \beta = \frac{\sin^2 \beta + \cos^2 \beta}{\sin \beta \cos \beta} = \frac{1}{\sin \beta \cos \beta} \]

### Step 2: Substitute back into the equation
Now we substitute this back into our equation:
\[ 2 \sec 2\alpha = \frac{1}{\sin \beta \cos \beta} \]
This simplifies to:
\[ 2 \cdot \frac{1}{\cos 2\alpha} = \frac{1}{\sin \beta \cos \beta} \]

### Step 3: Cross-multiply
Cross-multiplying gives:
\[ 2 \sin \beta \cos \beta = \cos 2\alpha \]

### Step 4: Use the double angle identity
Using the double angle identity for sine, we know:
\[ \sin 2\beta = 2 \sin \beta \cos \beta \]
Thus, we can rewrite our equation as:
\[ \sin 2\beta = \cos 2\alpha \]

### Step 5: Relate angles
From the equation \(\sin 2\beta = \cos 2\alpha\), we can use the co-function identity:
\[ \cos 2\alpha = \sin\left(\frac{\pi}{2} - 2\alpha\right) \]
This implies:
\[ 2\beta = \frac{\pi}{2} - 2\alpha + n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: Solve for \(\alpha + \beta\)
From the above equation:
\[ 2\beta + 2\alpha = \frac{\pi}{2} + n\pi \]
Dividing everything by 2 gives:
\[ \beta + \alpha = \frac{\pi}{4} + \frac{n\pi}{2} \]
Thus, one of the values of \(\alpha + \beta\) is:
\[ \alpha + \beta = \frac{\pi}{4} \quad \text{(for } n = 0\text{)} \]

### Final Answer
One of the values of \(\alpha + \beta\) is \(\frac{\pi}{4}\).

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