For `0 lt x lt (pi)/(2)`, let `P_(mn)(x)=m log_(cos x) ( sin x)+ n log_(cos x)(cotx)` ,
where `m, n in {1, 2,...,9}`
[For example: `P_(29)(x)=2log_(cosx)(sinx)+9log_(cos x)( cot x)` and
`" " P_(77)(x)=7 log_(cos x)(sin x)+(7 log_(cos x) ( cot x) `]
On the basis of above information, answer the following questions :
If `P_(34)(x)=P_(22)(x)`, then the value of sin x is expressed as `((sqrt(q)-1)/(p))`, then (p+q) equals

Options

A. (a)3
B. (b)4
C. (c)7
D. (d)9

Answer

To solve the problem step by step, we start with the given expression for \( P_{mn}(x) \):

### Step 1: Write the expressions for \( P_{34}(x) \) and \( P_{22}(x) \)

Given:
\[ P_{34}(x) = 3 \log_{\cos x} (\sin x) + 4 \log_{\cos x} (\cot x) \]
\[ P_{22}(x) = 2 \log_{\cos x} (\sin x) + 2 \log_{\cos x} (\cot x) \]

### Step 2: Set the two expressions equal to each other

Since we know that \( P_{34}(x) = P_{22}(x) \), we can write:
\[ 3 \log_{\cos x} (\sin x) + 4 \log_{\cos x} (\cot x) = 2 \log_{\cos x} (\sin x) + 2 \log_{\cos x} (\cot x) \]

### Step 3: Rearrange the equation

Rearranging the equation gives:
\[ 3 \log_{\cos x} (\sin x) - 2 \log_{\cos x} (\sin x) + 4 \log_{\cos x} (\cot x) - 2 \log_{\cos x} (\cot x) = 0 \]
This simplifies to:
\[ \log_{\cos x} (\sin x) + 2 \log_{\cos x} (\cot x) = 0 \]

### Step 4: Use the properties of logarithms

Using the property of logarithms, we can combine the logs:
\[ \log_{\cos x} (\sin x) + \log_{\cos x} (\cot^2 x) = 0 \]
This can be rewritten as:
\[ \log_{\cos x} (\sin x \cdot \cot^2 x) = 0 \]

### Step 5: Exponentiate to eliminate the logarithm

Exponentiating both sides gives:
\[ \sin x \cdot \cot^2 x = 1 \]

### Step 6: Substitute \( \cot x \)

Recall that \( \cot x = \frac{\cos x}{\sin x} \), so substituting gives:
\[ \sin x \cdot \left(\frac{\cos^2 x}{\sin^2 x}\right) = 1 \]
This simplifies to:
\[ \frac{\cos^2 x}{\sin x} = 1 \]

### Step 7: Rearranging the equation

Rearranging gives:
\[ \cos^2 x = \sin x \]

### Step 8: Use the Pythagorean identity

Using the identity \( \cos^2 x = 1 - \sin^2 x \), we can substitute:
\[ 1 - \sin^2 x = \sin x \]

### Step 9: Rearranging the equation

Rearranging gives:
\[ \sin^2 x + \sin x - 1 = 0 \]

### Step 10: Solve the quadratic equation

Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 1, c = -1 \):
\[ \sin x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 11: Determine the valid solution

Since \( 0 < x < \frac{\pi}{2} \), we need \( \sin x \) to be positive. Thus, we take:
\[ \sin x = \frac{-1 + \sqrt{5}}{2} \]

### Step 12: Express in the required form

We can express \( \sin x \) as:
\[ \sin x = \frac{\sqrt{5} - 1}{2} \]
Here, \( p = 2 \) and \( q = 5 \).

### Step 13: Calculate \( p + q \)

Thus, \( p + q = 2 + 5 = 7 \).

### Final Answer:
\[ \boxed{7} \]