If `(4 cos^(2)9^(@)-3)(4 cos^(2)27^(@)-3) =tanK^(@)` , then K is equal to

Answer

To solve the equation \( (4 \cos^2 9^\circ - 3)(4 \cos^2 27^\circ - 3) = \tan K^\circ \), we will use the trigonometric identity for \( \cos 3\theta \).

### Step-by-Step Solution:

1. **Use the identity for \( \cos 3\theta \)**:
The identity states that:
\[ \cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta \]
Rearranging gives us:
\[ 4 \cos^2 \theta - 3 = \frac{\cos 3\theta}{\cos \theta} \]

2. **Apply the identity to \( 9^\circ \)**:
For \( \theta = 9^\circ \):
\[ 4 \cos^2 9^\circ - 3 = \frac{\cos 27^\circ}{\cos 9^\circ} \]

3. **Apply the identity to \( 27^\circ \)**:
For \( \theta = 27^\circ \):
\[ 4 \cos^2 27^\circ - 3 = \frac{\cos 81^\circ}{\cos 27^\circ} \]

4. **Substituting back into the original equation**:
Now, substitute these results back into the left-hand side of the original equation:
\[ (4 \cos^2 9^\circ - 3)(4 \cos^2 27^\circ - 3) = \left(\frac{\cos 27^\circ}{\cos 9^\circ}\right) \left(\frac{\cos 81^\circ}{\cos 27^\circ}\right) \]

5. **Simplifying the expression**:
The \( \cos 27^\circ \) terms cancel out:
\[ = \frac{\cos 81^\circ}{\cos 9^\circ} \]

6. **Using the identity \( \cos 81^\circ \)**:
We know that:
\[ \cos 81^\circ = \sin 9^\circ \]
Therefore, we can rewrite the expression as:
\[ \frac{\sin 9^\circ}{\cos 9^\circ} = \tan 9^\circ \]

7. **Equating to \( \tan K^\circ \)**:
Now we have:
\[ \tan K^\circ = \tan 9^\circ \]

8. **Finding \( K \)**:
Since the tangent function is periodic, we can conclude:
\[ K = 9^\circ \]

### Final Answer:
Thus, \( K \) is equal to \( 9^\circ \).

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