Certainly! Let's solve the equation [tex]\(\tan^2(A) - \sin^2(A) = \tan^2(A) \cdot \sin^2(A)\)[/tex] step by step.
### Step 1: Understanding the Equation
We need to find angles [tex]\(A\)[/tex] that satisfy the equation:
[tex]\[ \tan^2(A) - \sin^2(A) = \tan^2(A) \sin^2(A) \][/tex]
### Step 2: Simplifying the Equation
First, let's take one of the most common approaches: substitution and trigonometric identities.
Recall that:
[tex]\[ \tan(A) = \frac{\sin(A)}{\cos(A)} \][/tex]
Thus:
[tex]\[ \tan^2(A) = \frac{\sin^2(A)}{\cos^2(A)} \][/tex]
Let's substitute [tex]\(\tan^2(A)\)[/tex] into the equation:
[tex]\[ \frac{\sin^2(A)}{\cos^2(A)} - \sin^2(A) = \frac{\sin^2(A)}{\cos^2(A)} \cdot \sin^2(A) \][/tex]
### Step 3: Getting a Common Denominator
To combine terms on the left-hand side, we need a common denominator:
[tex]\[ \frac{\sin^2(A) - \sin^2(A)\cos^2(A)}{\cos^2(A)} = \frac{\sin^2(A)}{\cos^2(A)} \cdot \sin^2(A) \][/tex]
[tex]\[ \frac{\sin^2(A)(1 - \cos^2(A))}{\cos^2(A)} = \frac{\sin^2(A) \cdot \sin^2(A)}{\cos^2(A)} \][/tex]
### Step 4: Using Trigonometric Identity
We know from Pythagoras’ theorem that:
[tex]\[ 1 - \cos^2(A) = \sin^2(A) \][/tex]
So, we substitute [tex]\(1 - \cos^2(A)\)[/tex] with [tex]\(\sin^2(A)\)[/tex]:
[tex]\[ \frac{\sin^4(A)}{\cos^2(A)} = \frac{\sin^4(A)}{\cos^2(A)} \][/tex]
### Step 5: Verify the Equation
We now have:
[tex]\[ \frac{\sin^4(A)}{\cos^2(A)} = \frac{\sin^4(A)}{\cos^2(A)} \][/tex]
This equation holds true as both sides are equal, confirming that our approach is correct.
### Step 6: Finding Solutions
To find the specific angles, we consider the cases [tex]\( \sin(A) = 0\)[/tex] and [tex]\( \tan(A) = 0 \)[/tex]. Solving for [tex]\(A\)[/tex] gives us:
1. [tex]\(\sin(A) = 0\)[/tex] yields [tex]\(A = n\pi\)[/tex], where [tex]\(n\)[/tex] is an integer.
2. Since [tex]\(\tan(A) = \frac{\sin(A)}{\cos(A)}\)[/tex], and knowing [tex]\(\sin(A) = 0\)[/tex] also results in [tex]\(\tan(A) = 0\)[/tex].
Thus, the general solutions to the trigonometric equation are:
[tex]\[ A = 0, \pm\pi, \pm2\pi, \ldots \][/tex]
### Final Answer
The specific solutions within a principal cycle (usually one cycle of [tex]\(2\pi\)[/tex]) can be enumerated as:
[tex]\[ A = 0, \pi, -\pi, 2\pi \][/tex]
This means the angles satisfying the given equation are:
[tex]\[ \boxed{0, \pi, -\pi, 2\pi} \][/tex]
These solutions fall within the principal interval and correctly solve the original trigonometric equation.
