Question

If \[ \sec\theta+\tan\theta=a \] then \[ \sin\theta= \]

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

Hint:

Use \((\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1\) to simplify quickly.

Answer 1

The Correct Option is A

Given: \[ \sec\theta+\tan\theta=a \] Using identity: \[ (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1 \] So: \[ \sec\theta-\tan\theta=\frac1a \] Add: \[ 2\sec\theta=a+\frac1a=\frac{a^2+1}{a} \] \[ \sec\theta=\frac{a^2+1}{2a} \] Subtract: \[ 2\tan\theta=a-\frac1a=\frac{a^2-1}{a} \] \[ \tan\theta=\frac{a^2-1}{2a} \] Now: \[ \sin\theta=\frac{\tan\theta}{\sec\theta} =\frac{\frac{a^2-1}{2a}}{\frac{a^2+1}{2a}} =\frac{a^2-1}{a^2+1} \] Thus, \[ \boxed{\frac{a^2-1}{a^2+1}} \]