Given expression to prove:
[tex]\[
\cos \theta \sqrt{1 + \cot^2 \theta} = \sqrt{\operatorname{cosec}^2 \theta} - 1
\][/tex]
### Step-by-Step Solution
1. Express [tex]\(\cot \theta\)[/tex] and [tex]\(\operatorname{cosec} \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[
\cot \theta = \frac{\cos \theta}{\sin \theta}
\][/tex]
[tex]\[
\operatorname{cosec} \theta = \frac{1}{\sin \theta}
\][/tex]
2. Substitute these definitions into the given expression:
[tex]\[
\cos \theta \sqrt{1 + \left( \frac{\cos \theta}{\sin \theta} \right)^2} = \sqrt{\frac{1}{\sin^2 \theta}} - 1
\][/tex]
3. Simplify the [tex]\(\sqrt{1 + \cot^2 \theta}\)[/tex] term:
[tex]\[
1 + \cot^2 \theta = 1 + \left( \frac{\cos \theta}{\sin \theta} \right)^2
= 1 + \frac{\cos^2 \theta}{\sin^2 \theta}
= \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta}
\][/tex]
Using the Pythagorean identity [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex]:
[tex]\[
= \frac{1}{\sin^2 \theta}
\][/tex]
4. Taking the square root of both sides:
[tex]\[
\sqrt{1 + \cot^2 \theta} = \sqrt{\frac{1}{\sin^2 \theta}} = \frac{1}{\sin \theta}
\][/tex]
5. Substitute back into the left-hand side of the original equation:
[tex]\[
\cos \theta \cdot \frac{1}{\sin \theta} = \frac{\cos \theta}{\sin \theta}
\][/tex]
6. Simplify the right-hand side of the original equation:
[tex]\[
\sqrt{\operatorname{cosec}^2 \theta} - 1 = \sqrt{\left( \frac{1}{\sin \theta} \right)^2} - 1 = \frac{1}{\sin \theta} - 1
\][/tex]
So now we have:
[tex]\[
\frac{\cos \theta}{\sin \theta} = \frac{1}{\sin \theta} - 1
\][/tex]
To show equivalency, let's compare the simplified forms:
7. Rewrite the simplified left-hand side:
[tex]\[
\frac{\cos \theta}{\sin \theta} = \cot \theta
\][/tex]
8. Rewrite the right-hand side:
[tex]\[
\operatorname{cosec} \theta - 1 = \frac{1}{\sin \theta} - 1
\][/tex]
Upon closer inspection, we observe that the left-hand side [tex]\(\cot \theta\)[/tex] and the adjusted right-hand side [tex]\(\frac{1}{\sin \theta} - 1\)[/tex] are not equal directly. This discrepancy suggests that the provided Python result, indicating [tex]\(None\)[/tex] as equivalence, confirms a deeper inspection of equivalency in trigonometric expressions.
Therefore, based on the provided simplification and reasoning:
[tex]\[
\cos \theta \sqrt{1 + \cot^2 \theta} \neq \sqrt{\operatorname{cosec}^2 \theta} - 1
\][/tex]
Hence, the given trigonometric identity does not hold true as shown by simplification steps and final comparison.
