Certainly! Let's solve the given trigonometric equation step-by-step:
[tex]\[ \cos^2 \theta + \cos^2 \theta \cdot \cot^2 \theta = \cot^2 \theta \][/tex]
First, we recall that the cotangent function can be expressed in terms of sine and cosine:
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
Thus,
[tex]\[ \cot^2 \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
Now, let's substitute [tex]\(\cot^2 \theta\)[/tex] in the original equation:
[tex]\[ \cos^2 \theta + \cos^2 \theta \cdot \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
Simplify the expression:
[tex]\[ \cos^2 \theta + \frac{\cos^4 \theta}{\sin^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta} \][/tex]
To clear the fraction, multiply the entire equation by [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \sin^2 \theta \cdot \cos^2 \theta + \cos^4 \theta = \cos^2 \theta \][/tex]
Next, use the Pythagorean identity [tex]\( \sin^2 \theta + \cos^2 \theta = 1 \)[/tex]:
[tex]\[ (1 - \cos^2 \theta) \cdot \cos^2 \theta + \cos^4 \theta = \cos^2 \theta \][/tex]
Expand and simplify the left-hand side:
[tex]\[ \cos^2 \theta - \cos^4 \theta + \cos^4 \theta = \cos^2 \theta \][/tex]
This simplifies to:
[tex]\[ \cos^2 \theta = \cos^2 \theta \][/tex]
Since this equation is always true for any [tex]\(\theta\)[/tex] where [tex]\(\sin \theta \ne 0\)[/tex], we have confirmed that the given trigonometric identity is valid for all such [tex]\(\theta\)[/tex]. Hence, the original equation holds true universally:
[tex]\[ \cos^2 \theta + \cos^2 \theta \cdot \cot^2 \theta = \cot^2 \theta \][/tex]
Thus, the result is [tex]\(\boxed{\text{True}}\)[/tex].
