If : `p = cot 20^(@),"then": (tan 160^(@) - tan 110^(@))/(1 +tan 160^(@) * tan 110^(@))=` A)`(p^(2)-1)/(2p)` B)`(p^(2)+1)/(2p)` C)`(1-p^(2))/(2p)` D)`(2p)/(1 + p^(2))`

Options

A. `(p^(2)-1)/(2p)`
B. `(p^(2)+1)/(2p)`
C. `(1-p^(2))/(2p)`
D. `(2p)/(1 + p^(2))`

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{\tan 160^\circ - \tan 110^\circ}{1 + \tan 160^\circ \tan 110^\circ} \]

Given that \( p = \cot 20^\circ \).

### Step-by-Step Solution:

1. **Use the Tangent Difference Formula**:
The formula for the tangent of the difference of two angles is:
\[ \tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \]
Here, let \( x = 160^\circ \) and \( y = 110^\circ \). Thus, we can rewrite the expression as:
\[ \tan(160^\circ - 110^\circ) = \tan 50^\circ \]

2. **Express \(\tan 50^\circ\)**:
We can express \(\tan 50^\circ\) in terms of cotangent:
\[ \tan 50^\circ = \tan(90^\circ - 40^\circ) = \cot 40^\circ \]

3. **Relate \(\cot 40^\circ\) to \(\cot 20^\circ\)**:
We know that:
\[ \cot 40^\circ = \cot(2 \times 20^\circ) = \frac{\cot^2 20^\circ - 1}{2 \cot 20^\circ} \]
Using the identity \(\cot(2\theta) = \frac{\cot^2 \theta - 1}{2 \cot \theta}\), we substitute \(\cot 20^\circ\) with \(p\):
\[ \cot 40^\circ = \frac{p^2 - 1}{2p} \]

4. **Final Result**:
Therefore, we have:
\[ \frac{\tan 160^\circ - \tan 110^\circ}{1 + \tan 160^\circ \tan 110^\circ} = \tan 50^\circ = \cot 40^\circ = \frac{p^2 - 1}{2p} \]

5. **Conclusion**:
The expression simplifies to:
\[ \frac{p^2 - 1}{2p} \]
Thus, the correct answer is option A: \((p^2 - 1)/(2p)\).