If \(\frac{\cos \theta}{1 + \sin \theta} = \frac{1}{a}\), then prove that \(\frac{a^2 - 1}{a^2 + 1} = \sin \theta\).
Proof:
\(\frac{1}{a} = \frac{\cos \theta}{1 + \sin \theta}\)
\(a=\)
\(a^2-1=\)
\(a^2+1=\)
From (1) and (2), \(\frac{a^2 - 1}{a^2 + 1} = \sin \theta\).
Answer variants:
\( \frac{\sin^2 \theta + 2 \sin \theta + 1 }{\cos^2 \theta} - 1\)
\( \frac{\cos \theta}{1 - \sin \theta}\)
\( \frac{1 + \sin \theta}{\cos \theta}\)
\( \frac{\sin^2 \theta + 2 \sin \theta - 1 }{\cos^2 \theta} +1\)
\( \frac{1 + 2 \sin \theta + \sin^2 \theta }{\cos^2 \theta} + 1\)