If the solution of the equation `log_(cosx) cotx+4log_(sinx) tanx=1, x in (0,pi/2)`,is `sin^(-1)((alpha+sqrt(beta))/(2))`,where `alpha,beta` are integers,then `alpha+beta` is equal to :

Options

A. `6`
B. `4`
C. `5`
D. `3`

Answer

To solve the equation \( \log_{\cos x} \cot x + 4 \log_{\sin x} \tan x = 1 \) for \( x \in (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Rewrite the logarithmic expressions
We start by rewriting the logarithmic expressions in terms of natural logarithms:

\[ \log_{\cos x} \cot x = \frac{\log \cot x}{\log \cos x} \quad \text{and} \quad \log_{\sin x} \tan x = \frac{\log \tan x}{\log \sin x} \]

Thus, the equation becomes:

\[ \frac{\log \cot x}{\log \cos x} + 4 \cdot \frac{\log \tan x}{\log \sin x} = 1 \]

### Step 2: Express cotangent and tangent
Recall that:

\[ \cot x = \frac{\cos x}{\sin x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \]

So we can rewrite the logarithms:

\[ \log \cot x = \log \cos x - \log \sin x \quad \text{and} \quad \log \tan x = \log \sin x - \log \cos x \]

### Step 3: Substitute back into the equation
Substituting these into our equation gives:

\[ \frac{\log \cos x - \log \sin x}{\log \cos x} + 4 \cdot \frac{\log \sin x - \log \cos x}{\log \sin x} = 1 \]

### Step 4: Simplify the equation
This simplifies to:

\[ 1 - \frac{\log \sin x}{\log \cos x} + 4 \left(1 - \frac{\log \cos x}{\log \sin x}\right) = 1 \]

Rearranging gives:

\[ -\frac{\log \sin x}{\log \cos x} + 4 - 4\frac{\log \cos x}{\log \sin x} = 0 \]

### Step 5: Let \( a = \log \sin x \) and \( b = \log \cos x \)
Let \( a = \log \sin x \) and \( b = \log \cos x \). The equation becomes:

\[ -\frac{a}{b} + 4 - 4\frac{b}{a} = 0 \]

Multiplying through by \( ab \) gives:

\[ -a^2 + 4ab - 4b^2 = 0 \]

### Step 6: Rearranging the quadratic equation
Rearranging gives:

\[ a^2 - 4ab + 4b^2 = 0 \]

This can be factored as:

\[ (a - 2b)^2 = 0 \]

### Step 7: Solve for \( a \)
Thus, we have:

\[ a - 2b = 0 \implies a = 2b \]

### Step 8: Substitute back to find \( x \)
Substituting back gives:

\[ \log \sin x = 2 \log \cos x \implies \sin x = \cos^2 x \]

### Step 9: Use the identity \( \sin x = 1 - \cos^2 x \)
Using the identity \( \sin^2 x + \cos^2 x = 1 \):

\[ \sin x = 1 - \sin^2 x \implies \sin^2 x + \sin x - 1 = 0 \]

### Step 10: Solve the quadratic equation
Using the quadratic formula:

\[ \sin x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

Since \( x \in (0, \frac{\pi}{2}) \), we take the positive root:

\[ \sin x = \frac{-1 + \sqrt{5}}{2} \]

### Step 11: Express in the required form
The solution is of the form \( \sin^{-1}\left(\frac{\alpha + \sqrt{\beta}}{2}\right) \), where \( \alpha = -1 \) and \( \beta = 5 \).

### Step 12: Calculate \( \alpha + \beta \)
Thus, we find:

\[ \alpha + \beta = -1 + 5 = 4 \]

### Final Answer
The final answer is \( \boxed{4} \).