Certainly! Let's solve the given equation step-by-step to show that [tex]\((\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{1 + \sin \theta}\)[/tex].
1. Express [tex]\(\sec \theta\)[/tex] and [tex]\(\tan \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[
\sec \theta = \frac{1}{\cos \theta}
\][/tex]
[tex]\[
\tan \theta = \frac{\sin \theta}{\cos \theta}
\][/tex]
2. Rewrite [tex]\(\sec \theta - \tan \theta\)[/tex] using the identities:
[tex]\[
\sec \theta - \tan \theta = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \frac{1 - \sin \theta}{\cos \theta}
\][/tex]
3. Square the expression:
[tex]\[
(\sec \theta - \tan \theta)^2 = \left( \frac{1 - \sin \theta}{\cos \theta} \right)^2 = \frac{(1 - \sin \theta)^2}{\cos^2 \theta}
\][/tex]
4. Use the Pythagorean identity [tex]\(\cos^2 \theta = 1 - \sin^2 \theta\)[/tex]:
[tex]\[
(\sec \theta - \tan \theta)^2 = \frac{(1 - \sin \theta)^2}{1 - \sin^2 \theta}
\][/tex]
5. Rewrite the denominator [tex]\((1 - \sin \theta)(1 + \sin \theta)\)[/tex]:
[tex]\[
(\sec \theta - \tan \theta)^2 = \frac{(1 - \sin \theta)^2}{(1 - \sin \theta)(1 + \sin \theta)}
\][/tex]
6. Simplify the fraction:
[tex]\[
(\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{1 + \sin \theta}
\][/tex]
This calculation confirms the given equation:
[tex]\[
(\sec \theta - \tan \theta)^2 = \frac{1 - \sin \theta}{1 + \sin \theta}
\][/tex]
Thus, the equation holds true for all [tex]\(\theta\)[/tex] satisfying the given trigonometric identities.
