(a) Show that \(\frac{\sin \theta}{1 - \sin \theta} - \frac{\sin \theta}{1 + \sin \theta} \equiv 2 \tan^2 \theta\).
(b) Hence solve the equation \(\frac{\sin \theta}{1 - \sin \theta} - \frac{\sin \theta}{1 + \sin \theta} = 8\), for \(0^\circ < \theta < 180^\circ\).
Solution
(a) Start by putting the expression over a single common denominator:
\(\frac{\sin \theta (1 + \sin \theta) - \sin \theta (1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)}\)
This simplifies to:
\(\frac{\sin \theta + \sin^2 \theta - \sin \theta + \sin^2 \theta}{1 - \sin^2 \theta}\)
\(= \frac{2 \sin^2 \theta}{\cos^2 \theta}\)
\(= 2 \tan^2 \theta\)
(b) From part (a), we have:
\(2 \tan^2 \theta = 8\)
\(\tan^2 \theta = 4\)
\(\tan \theta = \pm 2\)
Solving for \(\theta\) in the range \(0^\circ < \theta < 180^\circ\), we find:
\(\theta = 63.4^\circ, \ 116.6^\circ\)
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