If `cos9alpha=sinalpha` and `9alpha lt 90^(@)`, then the value of `tan5alpha` is

Options

A. `1/sqrt(3)`
B. `sqrt(3)`
C. 1
D. 0

Explanation

To solve the problem where \( \cos(9\alpha) = \sin(\alpha) \) and \( 9\alpha < 90^\circ \), we will follow these steps:

### Step 1: Use the identity for sine
We know that:
\[ \sin(\alpha) = \cos(90^\circ - \alpha) \]
Thus, we can rewrite the equation:
\[ \cos(9\alpha) = \cos(90^\circ - \alpha) \]

### Step 2: Set the angles equal
Since the cosine function is equal for two angles, we can set the angles equal to each other:
\[ 9\alpha = 90^\circ - \alpha \]

### Step 3: Solve for \( \alpha \)
Now, we will solve for \( \alpha \):
\[ 9\alpha + \alpha = 90^\circ \]
\[ 10\alpha = 90^\circ \]
\[ \alpha = \frac{90^\circ}{10} = 9^\circ \]

### Step 4: Find \( \tan(5\alpha) \)
Now that we have \( \alpha \), we can find \( \tan(5\alpha) \):
\[ 5\alpha = 5 \times 9^\circ = 45^\circ \]
Thus:
\[ \tan(5\alpha) = \tan(45^\circ) \]

### Step 5: Evaluate \( \tan(45^\circ) \)
We know that:
\[ \tan(45^\circ) = 1 \]

### Final Answer
Therefore, the value of \( \tan(5\alpha) \) is:
\[ \boxed{1} \]

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