If `tan2theta.tan3theta=1` , where `0^(@)lt theta lt90^(@)` then the value of `theta` is

Options

A. `22(1)/(2)@`
B. `18^(@)`
C. `24^(@)`
D. `30^(@)`

Answer

To solve the equation \( \tan(2\theta) \cdot \tan(3\theta) = 1 \) for \( 0^\circ < \theta < 90^\circ \), we can follow these steps:

### Step 1: Rewrite the equation
Given:
\[ \tan(2\theta) \cdot \tan(3\theta) = 1 \]
We can express \( \tan(3\theta) \) in terms of \( \tan(2\theta) \):
\[ \tan(3\theta) = \frac{1}{\tan(2\theta)} \]

### Step 2: Use the identity for tangent
Using the identity \( \tan(90^\circ - x) = \cot(x) \), we can rewrite \( \tan(3\theta) \):
\[ \tan(3\theta) = \tan(90^\circ - 2\theta) \]
This gives us:
\[ \tan(3\theta) = \cot(2\theta) \]

### Step 3: Set the equations equal
From the previous steps, we have:
\[ \frac{1}{\tan(2\theta)} = \cot(2\theta) \]
This means:
\[ \tan(2\theta) = \tan(90^\circ - 3\theta) \]

### Step 4: Set the angles equal
Since the tangents are equal, we can set the angles equal to each other:
\[ 2\theta = 90^\circ - 3\theta \]

### Step 5: Solve for \( \theta \)
Rearranging the equation:
\[ 2\theta + 3\theta = 90^\circ \]
\[ 5\theta = 90^\circ \]
\[ \theta = \frac{90^\circ}{5} = 18^\circ \]

### Final Answer
Thus, the value of \( \theta \) is:
\[ \theta = 18^\circ \]

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