Question

which expression is equivalent to the trigonometric expression \frac{1 - \sin^{2}(\theta)}{\cos(\theta)}?

Explanation:

Step1: Use Pythagorean identity

Recall the Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta = 1\), so \(1-\sin^{2}\theta=\cos^{2}\theta\).
The given expression \(\frac{1 - \sin^{2}(\theta)}{\cos(\theta)}\) becomes \(\frac{\cos^{2}\theta}{\cos\theta}\).

Step2: Simplify the expression

When we have \(\frac{\cos^{2}\theta}{\cos\theta}\), using the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\) (\(a
eq0\)), here \(a = \cos\theta\), \(m = 2\), \(n=1\). So \(\frac{\cos^{2}\theta}{\cos\theta}=\cos\theta\).

Answer:

\(\cos(\theta)\)