Show that \(\frac{\sin \theta + 2 \cos \theta}{\cos \theta - 2 \sin \theta} - \frac{\sin \theta - 2 \cos \theta}{\cos \theta + 2 \sin \theta} \equiv \frac{4}{5 \cos^2 \theta - 4}\).
Solution
Start by obtaining a common denominator for the two fractions:
\(\frac{(\sin \theta + 2 \cos \theta)(\cos \theta + 2 \sin \theta) - (\sin \theta - 2 \cos \theta)(\cos \theta - 2 \sin \theta)}{(\cos \theta - 2 \sin \theta)(\cos \theta + 2 \sin \theta)}\)
Expand the numerators:
\(5 \sin \theta \cos \theta + 2 \sin^2 \theta + 2 \cos^2 \theta - (5 \sin \theta \cos \theta - 2 \sin^2 \theta - 2 \cos^2 \theta)\)
Simplify the expression:
\(\frac{4(\cos^2 \theta + \sin^2 \theta)}{\cos^2 \theta - 4 \sin^2 \theta}\)
Use the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\(\frac{4}{\cos^2 \theta - 4 \sin^2 \theta}\)
Recognize that \(\cos^2 \theta - 4 \sin^2 \theta = 5 \cos^2 \theta - 4\) by using \(\cos^2 \theta + \sin^2 \theta = 1\) again:
\(\frac{4}{5 \cos^2 \theta - 4}\)
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