Step 1: Understanding the Concept:
We are given an equation in terms of \( \sec A \) and \( \tan A \). We can use the fundamental identity \( \sec^2 A - \tan^2 A = 1 \) to relate these terms.
: Key Formula or Approach:
1. \( \sec A - \tan A = -a \).
2. \( (\sec A - \tan A)(\sec A + \tan A) = 1 \).
Step 2: Detailed Explanation:
From the given equation:
\[ \sec A - \tan A = -a \quad \dots (1) \]
We know \( \sec^2 A - \tan^2 A = 1 \), so:
\[ \sec A + \tan A = \frac{1}{\sec A - \tan A} = -\frac{1}{a} \quad \dots (2) \]
Subtracting equation (1) from equation (2):
\[ (\sec A + \tan A) - (\sec A - \tan A) = -\frac{1}{a} - (-a) \]
\[ 2 \tan A = a - \frac{1}{a} = \frac{a^2 - 1}{a} \implies \tan A = \frac{a^2 - 1}{2a} \]
Using the triangle relation: \( \tan A = \frac{P}{B} = \frac{a^2-1}{2a} \).
Hypotenuse \( H = \sqrt{P^2 + B^2} = \sqrt{(a^2-1)^2 + (2a)^2} = \sqrt{a^4 - 2a^2 + 1 + 4a^2} = \sqrt{(a^2+1)^2} = a^2+1 \).
Now find \( \sin A = \frac{P}{H} \).
Since \( \sec A - \tan A = -a \), if \( a \) is positive, \( \tan A \) must be more positive than \( \sec A \), indicating a specific quadrant.
Actually, from the relation \( \sec A - \tan A = \frac{1-\sin A}{\cos A} = -a \), squaring gives \( \frac{(1-\sin A)^2}{1-\sin^2 A} = a^2 \implies \frac{1-\sin A}{1+\sin A} = a^2 \).
\[ 1 - \sin A = a^2 + a^2 \sin A \]
\[ 1 - a^2 = \sin A(1 + a^2) \]
\[ \sin A = \frac{1 - a^2}{1 + a^2} \].
Step 3: Final Answer:
The value of \( \sin A \) is \( \frac{1 - a^2}{1 + a^2} \).