A dip needle lies initially in the magnetic merdian when it shows an angle of dip `theta` at a place. The dip circle is rotated through an angle x in the horizontal plane and then it shows an angle of dip `theta^(')`. Then `tantheta^(')/tantheta` is

Answer

To solve the problem, we need to analyze the relationship between the angles of dip before and after the dip circle is rotated.

### Step-by-Step Solution:

1. **Understanding the Dip Angle**:
The angle of dip (θ) is defined as the angle made by the Earth's magnetic field with the horizontal plane. It can be expressed in terms of the vertical and horizontal components of the magnetic field:
\[ \tan(\theta) = \frac{B_v}{B_h} \]
where \(B_v\) is the vertical component and \(B_h\) is the horizontal component of the magnetic field.

2. **Rotation of the Dip Circle**:
When the dip circle is rotated through an angle \(x\) in the horizontal plane, the vertical component of the magnetic field remains unchanged (\(B_v\)), but the horizontal component changes. The new horizontal component \(B_h'\) can be expressed as:
\[ B_h' = B_h \cos(x) \]

3. **New Angle of Dip**:
After the rotation, the new angle of dip (θ') can be expressed as:
\[ \tan(\theta') = \frac{B_v}{B_h'} = \frac{B_v}{B_h \cos(x)} \]

4. **Finding the Ratio**:
Now, we want to find the ratio \(\frac{\tan(\theta')}{\tan(\theta)}\):
\[ \frac{\tan(\theta')}{\tan(\theta)} = \frac{\frac{B_v}{B_h \cos(x)}}{\frac{B_v}{B_h}} = \frac{B_v}{B_h \cos(x)} \cdot \frac{B_h}{B_v} \]
Here, \(B_v\) cancels out, and we are left with:
\[ \frac{\tan(\theta')}{\tan(\theta)} = \frac{1}{\cos(x)} \]

5. **Final Result**:
Hence, we can conclude that:
\[ \frac{\tan(\theta')}{\tan(\theta)} = \sec(x) \]

### Final Answer:
\[ \frac{\tan(\theta')}{\tan(\theta)} = \frac{1}{\cos(x)} \]