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The question in the image asks to prove the following identity:

Given \( \tan 35^\circ = a \), prove that:
\[
\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ \tan 125^\circ} = \frac{1 - a^2}{2a}
\]

### Proof:

1. **Use of Angle Values**:
   We know that:
   \[
   \tan 145^\circ = \tan (180^\circ - 35^\circ) = -\tan 35^\circ = -a
   \]
   Similarly:
   \[
   \tan 125^\circ = \tan (180^\circ - 55^\circ) = -\tan 55^\circ
   \]
   Now, using the complementary angle identity \( \tan 55^\circ = \cot 35^\circ \), and we know:
   \[
   \cot \theta = \frac{1}{\tan \theta}
   \]
   Therefore:
   \[
   \tan 125^\circ = -\cot 35^\circ = -\frac{1}{a}
   \]

2. **Substitute into the Given Expression**:
   Now substitute these values into the left-hand side:
   \[
   \frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ \tan 125^\circ} = \frac{-a - \left(-\frac{1}{a}\right)}{1 + \left(-a\right)\left(-\frac{1}{a}\right)}
   \]
   Simplifying the numerator:
   \[
   -a + \frac{1}{a} = \frac{-a^2 + 1}{a}
   \]
   Simplifying the denominator:
   \[
   1 + a \cdot \frac{1}{a} = 1 + 1 = 2
   \]
   Thus, the expression becomes:
   \[
   \frac{\frac{-a^2 + 1}{a}}{2} = \frac{-a^2 + 1}{2a} = \frac{1 - a^2}{2a}
   \]

Thus, we have proved that:
\[
\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ \tan 125^\circ} = \frac{1 - a^2}{2a}
\]

### Do you want further details or have any questions?

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Here are 5 related questions to expand your understanding:
1. How does the tangent function behave for angles greater than \(90^\circ\)?
2. Can this identity be generalized for other angles?
3. How is the complementary angle identity \( \tan(90^\circ - \theta) = \cot(\theta) \) derived?
4. What other trigonometric identities involve products of tangent functions?
5. How can the tangent addition formula be used to simplify complex trigonometric expressions?

### Tip:
For trigonometric proofs, knowing identities like the tangent addition formula and complementary angle properties is key to simplifying and manipulating expressions efficiently.

If tan 35° = a, prove that (tan 145° - tan 125°) / (1 + tan 145° tan 125°) = (1 - a²) / 2a.

In this problem, we prove a trigonometric identity where tan 35° = a, and show that (tan 145° - tan 125°) / (1 + tan 145° tan 125°) simplifies to (1 - a²) / 2a. Key concepts include the tangent function, complementary angles, and trigonometric identities. Suitable for Grades 10-12.

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