Answer :

Certainly! Let’s break down the given trigonometric identity step-by-step: [tex]\(\cos^2(\theta) \cdot \tan^2(\theta) = \sin^2(\theta)\)[/tex].

1. Understand the Identity:
The given identity is [tex]\(\cos^2(\theta) \cdot \tan^2(\theta) = \sin^2(\theta)\)[/tex]. We need to verify if this is true.

2. Express [tex]\(\tan(\theta)\)[/tex] in Terms of Sine and Cosine:
Recall that the tangent of an angle [tex]\(\theta\)[/tex] is given by the ratio of the sine and the cosine of that angle.
[tex]\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \][/tex]
Therefore, squaring both sides, we get:
[tex]\[ \tan^2(\theta) = \left( \frac{\sin(\theta)}{\cos(\theta)} \right)^2 = \frac{\sin^2(\theta)}{\cos^2(\theta)} \][/tex]

3. Substitute [tex]\(\tan^2(\theta)\)[/tex] in the Original Identity:
Replace [tex]\(\tan^2(\theta)\)[/tex] in the identity [tex]\(\cos^2(\theta) \cdot \tan^2(\theta) = \sin^2(\theta)\)[/tex] with [tex]\(\frac{\sin^2(\theta)}{\cos^2(\theta)}\)[/tex]:
[tex]\[ \cos^2(\theta) \cdot \left(\frac{\sin^2(\theta)}{\cos^2(\theta)}\right) = \sin^2(\theta) \][/tex]

4. Simplify the Expression:
On the left-hand side of the equation, the [tex]\(\cos^2(\theta)\)[/tex] in the numerator and the [tex]\(\cos^2(\theta)\)[/tex] in the denominator cancel each other out:
[tex]\[ \left(\frac{\cos^2(\theta)}{\cos^2(\theta)}\right) \cdot \sin^2(\theta) = \sin^2(\theta) \][/tex]
Since [tex]\(\cos^2(\theta) / \cos^2(\theta) = 1\)[/tex] (assuming [tex]\(\cos(\theta)\)[/tex] is not zero), the expression further simplifies to:
[tex]\[ 1 \cdot \sin^2(\theta) = \sin^2(\theta) \][/tex]
Hence, we have:
[tex]\[ \sin^2(\theta) = \sin^2(\theta) \][/tex]

5. Conclusion:
The left-hand side and the right-hand side of the equation are equal under the condition that [tex]\(\cos(\theta) \neq 0\)[/tex]. Thus, the given trigonometric identity [tex]\(\cos^2(\theta) \cdot \tan^2(\theta) = \sin^2(\theta)\)[/tex] is indeed valid for values of [tex]\(\theta\)[/tex] where [tex]\(\cos(\theta) \neq 0\)[/tex].

So, the identity [tex]\(\cos^2(\theta) \cdot \tan^2(\theta) = \sin^2(\theta)\)[/tex] holds true where [tex]\(\cos(\theta) \neq 0\)[/tex].