Certainly! Let's solve the given expression step-by-step:
### Given Expression:
[tex]\[ \tan^2 A \cdot \operatorname{cosec}^2 A - \cos 2\pi \][/tex]
First, let's break down each component of the given expression:
1. [tex]\(\tan^2 A\)[/tex]:
[tex]\[ \tan A = \frac{\sin A}{\cos A} \][/tex]
[tex]\[ \tan^2 A = \left( \frac{\sin A}{\cos A} \right)^2 = \frac{\sin^2 A}{\cos^2 A} \][/tex]
2. [tex]\(\operatorname{cosec}^2 A\)[/tex]:
[tex]\[ \operatorname{cosec} A = \frac{1}{\sin A} \][/tex]
[tex]\[ \operatorname{cosec}^2 A = \left( \frac{1}{\sin A} \right)^2 = \frac{1}{\sin^2 A} \][/tex]
3. Combine [tex]\(\tan^2 A\)[/tex] and [tex]\(\operatorname{cosec}^2 A\)[/tex]:
[tex]\[ \tan^2 A \cdot \operatorname{cosec}^2 A = \frac{\sin^2 A}{\cos^2 A} \cdot \frac{1}{\sin^2 A} = \frac{1}{\cos^2 A} = \sec^2 A \][/tex]
4. [tex]\(\cos 2\pi\)[/tex]:
[tex]\[ \cos 2\pi = 1 \][/tex]
This is because the cosine of [tex]\(2\pi\)[/tex] (which is a full rotation in radians) returns to the starting point on the unit circle, which is 1.
### Putting everything together:
Now, substitute these values back into the given expression:
[tex]\[ \tan^2 A \cdot \operatorname{cosec}^2 A - \cos 2\pi = \sec^2 A - 1 \][/tex]
According to trigonometric identities:
[tex]\[ \sec^2 A = 1 + \tan^2 A \][/tex]
So,
[tex]\[ \sec^2 A - 1 = \tan^2 A \][/tex]
This simplifies our expression to:
[tex]\[ \boxed{\tan^2 A - 1 + \tan^2 A / \sin^2 A} = \tan^2 A \cdot \operatorname{cosec}^2 A - \cos 2\pi \][/tex]
Finally, substituting the expression we found, we get the simplified result:
[tex]\[ \boxed{-1 + \frac{\tan^2 A}{\sin^2 A}} = \boxed{-1 + \tan^2 A \cdot \operatorname{cosec}^2 A} \][/tex]
Therefore, the expression simplifies to:
[tex]\[ \boxed{-1 + \tan^2(A) \cdot \text{cosec}^2(A)} \][/tex]
