Question

If \(\sec A - \tan A + a = 0\), then \(\sin A\) is equal to

• A. \(\frac{a^2 -1}{a^2 +1}\)
• B. \(\frac{1-a^2}{1+a^2}\)
• C. \(\frac{a^2 +1}{a^2 -1}\)
• D. \(\frac{1+a^2}{1-a^2}\)

Hint:

Use \((\sec A - \tan A)(\sec A + \tan A)=1\) to quickly relate expressions.

Answer 1

The Correct Option is B

Concept: \[ (\sec A - \tan A)(\sec A + \tan A)=1 \]

Step 1: Given condition.
\[ \sec A - \tan A = -a \Rightarrow \sec A + \tan A = -\frac{1}{a} \]

Step 2: Add and subtract.
\[ 2\sec A = -a - \frac{1}{a} \Rightarrow \sec A = -\frac{a^2+1}{2a} \] \[ 2\tan A = -a + \frac{1}{a} \Rightarrow \tan A = \frac{1-a^2}{2a} \]

Step 3: Find \(\sin A\).
\[ \sin A = \frac{\tan A}{\sec A} = \frac{\frac{1-a^2}{2a}}{-\frac{a^2+1}{2a}} = \frac{1-a^2}{1+a^2} \]