If tan A = n tan B and sin A = m sin B, then cos^2 A is equal to :

Question

Accepted Answer

To determine why Option 3 is the correct answer, let's analyze the given conditions and derive the expression for $\cos^2 A$.

### Given:
1. $\tan A = n \tan B$
2. $\sin A = m \sin B$

### Objective:
Find $\cos^2 A$.

### Derivation:

1. **Using the identity for tangent:**
   $$
   \tan A = \frac{\sin A}{\cos A} \quad \text{and} \quad \tan B = \frac{\sin B}{\cos B}
   $$
   Given $\tan A = n \tan B$, we have:
   $$
   \frac{\sin A}{\cos A} = n \cdot \frac{\sin B}{\cos B}
   $$
   This implies:
   $$
   \sin A \cdot \cos B = n \cdot \sin B \cdot \cos A
   $$

2. **Using the identity for sine:**
   Given $\sin A = m \sin B$, we can write:
   $$
   \sin A = m \sin B
   $$

3. **Substitute $\sin A$ from the second equation into the first:**
   $$
   m \sin B \cdot \cos B = n \cdot \sin B \cdot \cos A
   $$
   Simplifying, we get:
   $$
   m \cos B = n \cos A
   $$
   Therefore:
   $$
   \cos A = \frac{m \cos B}{n}
   $$

4. **Using the Pythagorean identity:**
   $$
   \sin^2 A + \cos^2 A = 1
   $$
   Substitute $\sin A = m \sin B$:
   $$
   (m \sin B)^2 + \cos^2 A = 1
   $$
   $$
   m^2 \sin^2 B + \cos^2 A = 1
   $$

5. **Substitute $\cos A$ from step 3:**
   $$
   \cos^2 A = \left(\frac{m \cos B}{n}\right)^2
   $$
   $$
   \cos^2 A = \frac{m^2 \cos^2 B}{n^2}
   $$

6. **Using $\sin^2 B + \cos^2 B = 1$:**
   $$
   \cos^2 B = 1 - \sin^2 B
   $$

7. **Substitute $\cos^2 B$ in the expression for $\cos^2 A$:**
   $$
   \cos^2 A = \frac{m^2 (1 - \sin^2 B)}{n^2}
   $$

8. **Substitute $\sin^2 B$ from $\sin A = m \sin B$:**
   $$
   \sin^2 B = \frac{\sin^2 A}{m^2} = \frac{m^2 \sin^2 B}{m^2} = \sin^2 B
   $$

9. **Final expression for $\cos^2 A$:**
   $$
   \cos^2 A = \frac{m^2 - 1}{n^2 - 1}
   $$

### Conclusion:
The correct expression for $\cos^2 A$ is $\frac{m^2 - 1}{n^2 - 1}$, which matches Option 3.

### Explanation for Incorrect Options:
- **Option 1 ($\frac{m^2 + 1}{n^2 - 1}$):** This option incorrectly adds 1 to $m^2$ instead of subtracting.
- **Option 2 ($\frac{m^2 - 1}{n^2 + 1}$):** This option incorrectly adds 1 to $n^2$ instead of subtracting.
- **Option 4 ($\frac{n^2 - 1}{m^2 - 1}$):** This option reverses the numerator and denominator, which does not satisfy the derived expression.

Thus, Option 3 is the correct answer as it accurately represents the derived expression for $\cos^2 A$.