If `tantheta=cot(30^(@)+theta)`, Find the value of `theta`.

Answer

To solve the equation \( \tan \theta = \cot(30^\circ + \theta) \), we can follow these steps:

### Step 1: Use the cotangent identity
We know that \( \cot x = \frac{1}{\tan x} \). Therefore, we can rewrite the equation as:
\[ \tan \theta = \frac{1}{\tan(30^\circ + \theta)} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ \tan \theta \cdot \tan(30^\circ + \theta) = 1 \]

### Step 3: Use the tangent addition formula
Using the tangent addition formula, we have:
\[ \tan(30^\circ + \theta) = \frac{\tan 30^\circ + \tan \theta}{1 - \tan 30^\circ \tan \theta} \]
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), we can substitute this into the equation:
\[ \tan \theta \cdot \left( \frac{\frac{1}{\sqrt{3}} + \tan \theta}{1 - \frac{1}{\sqrt{3}} \tan \theta} \right) = 1 \]

### Step 4: Simplify the equation
Multiplying both sides by the denominator:
\[ \tan \theta \left( \frac{1}{\sqrt{3}} + \tan \theta \right) = 1 - \frac{1}{\sqrt{3}} \tan \theta \]
Expanding gives:
\[ \frac{\tan \theta}{\sqrt{3}} + \tan^2 \theta = 1 - \frac{1}{\sqrt{3}} \tan \theta \]

### Step 5: Rearranging the equation
Rearranging the equation leads to:
\[ \tan^2 \theta + \left( \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} \right) \tan \theta - 1 = 0 \]
This simplifies to:
\[ \tan^2 \theta + \frac{2}{\sqrt{3}} \tan \theta - 1 = 0 \]

### Step 6: Solve the quadratic equation
Let \( x = \tan \theta \). The equation becomes:
\[ x^2 + \frac{2}{\sqrt{3}} x - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-\frac{2}{\sqrt{3}} \pm \sqrt{\left(\frac{2}{\sqrt{3}}\right)^2 + 4}}{2} \]
Calculating the discriminant:
\[ \left(\frac{2}{\sqrt{3}}\right)^2 + 4 = \frac{4}{3} + 4 = \frac{4}{3} + \frac{12}{3} = \frac{16}{3} \]
Thus:
\[ x = \frac{-\frac{2}{\sqrt{3}} \pm \frac{4}{\sqrt{3}}}{2} \]
This simplifies to:
\[ x = \frac{2}{\sqrt{3}} \text{ or } x = -\frac{3}{\sqrt{3}} = -\sqrt{3} \]

### Step 7: Find the angles
Now, we find \( \theta \):
1. For \( \tan \theta = \frac{2}{\sqrt{3}} \), we have \( \theta = 30^\circ \).
2. For \( \tan \theta = -\sqrt{3} \), we have \( \theta = 150^\circ \) or \( \theta = -30^\circ \).

### Conclusion
The principal solution is:
\[ \theta = 30^\circ \]