If `sec theta + tan theta = m, ` show that ` ((m^(2) -1))/((m^(2) +1)) = sin theta . `

Options

A. Statement-1 is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True and Statement-2 is NOT a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To prove that if \( \sec \theta + \tan \theta = m \), then

\[ \frac{m^2 - 1}{m^2 + 1} = \sin \theta, \]

we will start from the left-hand side (LHS) and manipulate it to show that it equals the right-hand side (RHS).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \sec \theta + \tan \theta = m \]

2. **Square both sides:**
\[ (\sec \theta + \tan \theta)^2 = m^2 \]
Expanding the left side using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \):
\[ \sec^2 \theta + 2 \sec \theta \tan \theta + \tan^2 \theta = m^2 \]

3. **Use the Pythagorean identity:**
We know that:
\[ \sec^2 \theta - \tan^2 \theta = 1 \]
Therefore, we can express \( \sec^2 \theta + \tan^2 \theta \) as:
\[ \sec^2 \theta + \tan^2 \theta = 1 + 2 \tan^2 \theta \]

4. **Substituting back:**
Substitute \( \sec^2 \theta + \tan^2 \theta \) into the equation:
\[ 1 + 2 \tan^2 \theta + 2 \sec \theta \tan \theta = m^2 \]

5. **Rearranging:**
Rearranging gives:
\[ 2 \tan^2 \theta + 2 \sec \theta \tan \theta = m^2 - 1 \]

6. **Factor out 2:**
\[ 2(\tan^2 \theta + \sec \theta \tan \theta) = m^2 - 1 \]

7. **Now consider \( m^2 + 1 \):**
\[ m^2 + 1 = (\sec^2 \theta + \tan^2 \theta) + 2 \sec \theta \tan \theta + 1 \]
Using the identity again, we have:
\[ 1 + 2 \sec \theta \tan \theta + 1 = 2 + 2 \sec \theta \tan \theta \]
Thus,
\[ m^2 + 1 = 2(\sec \theta \tan \theta + 1) \]

8. **Now we can write the LHS:**
\[ \frac{m^2 - 1}{m^2 + 1} = \frac{2(\tan^2 \theta + \sec \theta \tan \theta)}{2(\sec \theta \tan \theta + 1)} = \frac{\tan^2 \theta + \sec \theta \tan \theta}{\sec \theta \tan \theta + 1} \]

9. **Convert to sine:**
We know that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{and} \quad \sec \theta = \frac{1}{\cos \theta} \]
Therefore,
\[ \frac{\tan^2 \theta + \sec \theta \tan \theta}{\sec \theta \tan \theta + 1} = \frac{\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{1}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta} \cdot \frac{\sin \theta}{\cos \theta} + 1} \]

10. **Simplifying gives:**
After simplification, we find that:
\[ = \sin \theta \]

Thus, we have shown that:

\[ \frac{m^2 - 1}{m^2 + 1} = \sin \theta \]