If 3 sin A = 4 cos A , find the value of :
`tan^2A-sec^2A`

Answer

To solve the problem, we need to find the value of \( \tan^2 A - \sec^2 A \) given that \( 3 \sin A = 4 \cos A \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ 3 \sin A = 4 \cos A \]

2. **Rearrange the equation to find the ratio of sine to cosine:**
\[ \frac{\sin A}{\cos A} = \frac{4}{3} \]
This implies:
\[ \tan A = \frac{4}{3} \]

3. **Use the Pythagorean theorem to find the hypotenuse:**
- In a right triangle, let the opposite side (perpendicular) be 4 and the adjacent side (base) be 3.
- The hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{(4^2) + (3^2)} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

4. **Calculate \( \tan^2 A \):**
\[ \tan^2 A = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]

5. **Calculate \( \sec A \):**
- The secant function is defined as:
\[ \sec A = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{5}{3} \]

6. **Calculate \( \sec^2 A \):**
\[ \sec^2 A = \left(\frac{5}{3}\right)^2 = \frac{25}{9} \]

7. **Now substitute \( \tan^2 A \) and \( \sec^2 A \) into the expression \( \tan^2 A - \sec^2 A \):**
\[ \tan^2 A - \sec^2 A = \frac{16}{9} - \frac{25}{9} \]

8. **Combine the fractions:**
\[ \tan^2 A - \sec^2 A = \frac{16 - 25}{9} = \frac{-9}{9} = -1 \]

9. **Final answer:**
\[ \tan^2 A - \sec^2 A = -1 \]