To verify the trigonometric identity [tex]\((\operatorname{csc} \theta - \operatorname{cot} \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}\)[/tex], let's simplify both sides step by step and check if they are indeed equal.
### Step 1: Simplify the Left-Hand Side (LHS) The left-hand side of the equation is: [tex]\[
(\operatorname{csc} \theta - \operatorname{cot} \theta)^2
\][/tex]
First, recall the definitions of [tex]\(\operatorname{csc} \theta\)[/tex] and [tex]\(\operatorname{cot} \theta\)[/tex]: [tex]\[
\operatorname{csc} \theta = \frac{1}{\sin \theta} \quad \text{and} \quad \operatorname{cot} \theta = \frac{\cos \theta}{\sin \theta}
\][/tex]
Substituting these into the LHS: [tex]\[
(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta})^2
\][/tex]
Simplify the expression inside the square: [tex]\[
\frac{(1 - \cos \theta)^2}{\sin^2 \theta}
\][/tex]
### Step 2: Simplify the Right-Hand Side (RHS) The right-hand side of the equation is: [tex]\[
\frac{1 - \cos \theta}{1 + \cos \theta}
\][/tex]
### Step 3: Evaluate the Equality We need to check if the simplified LHS is equal to the simplified RHS. Recall an important trigonometric identity: [tex]\[
1 - \cos^2 \theta = \sin^2 \theta
\][/tex]
Using this, we can rewrite the simplified LHS in terms of [tex]\(\sin^2 \theta\)[/tex]: [tex]\[
\frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}
\][/tex]
Because [tex]\(1 - \cos \theta\)[/tex] is a common factor in both the numerator and numerator when squaring: [tex]\[
\frac{(1 - \cos \theta)^2}{\sin^2 \theta} = \frac{(1 - \cos \theta)(1 - \cos \theta)}{1 - \cos^2 \theta}
\][/tex]
At this point, it is evident that factoring [tex]\(1-\cos^2 \theta\)[/tex]: [tex]\[
\rightarrow \frac{1 - \cos \theta}{1 + \cos \theta}
\][/tex]
