Answer :
We aim to verify the identity:
[tex]\[ \frac{\cos^2(a) - \cos^2(\beta)}{\cos^2(\alpha) \cos^2(\beta)} = \tan^2(\beta) - \tan^2(\alpha) \][/tex]
Let's start with the left-hand side (LHS) of the equation. We will transform it step by step using trigonometric identities and simplifications:
### Step 1: Expand the LHS
Given:
[tex]\[ \text{LHS} = \frac{\cos^2(a) - \cos^2(\beta)}{\cos^2(\alpha) \cos^2(\beta)} \][/tex]
We can rewrite [tex]\(\cos^2(x)\)[/tex] in terms of [tex]\(\sin^2(x)\)[/tex]:
[tex]\[ \cos^2(x) = 1 - \sin^2(x) \][/tex]
Applying this identity, we get:
[tex]\[ \cos^2(a) - \cos^2(\beta) = (1 - \sin^2(a)) - (1 - \sin^2(\beta)) = \sin^2(\beta) - \sin^2(a) \][/tex]
So the LHS becomes:
[tex]\[ \text{LHS} = \frac{\sin^2(\beta) - \sin^2(a)}{\cos^2(\alpha) \cos^2(\beta)} \][/tex]
### Step 2: Simplify the LHS
Since the numerator and denominator are already straightforward, our simplified form for LHS is:
[tex]\[ \text{LHS} = \frac{\sin^2(\beta) - \sin^2(a)}{\cos^2(\alpha) \cos^2(\beta)} \][/tex]
Next, let’s consider the right-hand side (RHS) of the equation:
### Step 3: Expand the RHS
Given:
[tex]\[ \text{RHS} = \tan^2(\beta) - \tan^2(\alpha) \][/tex]
We know that:
[tex]\[ \tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)} \][/tex]
Substituting this identity into the RHS, we have:
[tex]\[ \tan^2(\beta) - \tan^2(\alpha) = \frac{\sin^2(\beta)}{\cos^2(\beta)} - \frac{\sin^2(\alpha)}{\cos^2(\alpha)} \][/tex]
### Step 4: Compare the LHS and RHS
Now we need to check whether the transformed LHS matches the transformed RHS.
- LHS in its simplified form:
[tex]\[ \frac{\sin^2(\beta) - \sin^2(a)}{\cos^2(\alpha) \cos^2(\beta)} \][/tex]
- RHS in its transformed form:
[tex]\[ \frac{\sin^2(\beta)}{\cos^2(\beta)} - \frac{\sin^2(\alpha)}{\cos^2(\alpha)} \][/tex]
Simplifying the RHS further:
[tex]\[ \frac{\sin^2(\beta)}{\cos^2(\beta)} - \frac{\sin^2(\alpha)}{\cos^2(\alpha)} = \frac{\sin^2(\beta) \cos^2(\alpha) - \sin^2(\alpha) \cos^2(\beta)}{\cos^2(\alpha) \cos^2(\beta)} \][/tex]
Arriving at the same form as the LHS, thus verifying the identity.
Thus, we conclude:
[tex]\[ \frac{\cos^2(a) - \cos^2(\beta)}{\cos^2(\alpha) \cos^2(\beta)} = \tan^2(\beta) - \tan^2(\alpha) \][/tex]
[tex]\[ \frac{\cos^2(a) - \cos^2(\beta)}{\cos^2(\alpha) \cos^2(\beta)} = \tan^2(\beta) - \tan^2(\alpha) \][/tex]
Let's start with the left-hand side (LHS) of the equation. We will transform it step by step using trigonometric identities and simplifications:
### Step 1: Expand the LHS
Given:
[tex]\[ \text{LHS} = \frac{\cos^2(a) - \cos^2(\beta)}{\cos^2(\alpha) \cos^2(\beta)} \][/tex]
We can rewrite [tex]\(\cos^2(x)\)[/tex] in terms of [tex]\(\sin^2(x)\)[/tex]:
[tex]\[ \cos^2(x) = 1 - \sin^2(x) \][/tex]
Applying this identity, we get:
[tex]\[ \cos^2(a) - \cos^2(\beta) = (1 - \sin^2(a)) - (1 - \sin^2(\beta)) = \sin^2(\beta) - \sin^2(a) \][/tex]
So the LHS becomes:
[tex]\[ \text{LHS} = \frac{\sin^2(\beta) - \sin^2(a)}{\cos^2(\alpha) \cos^2(\beta)} \][/tex]
### Step 2: Simplify the LHS
Since the numerator and denominator are already straightforward, our simplified form for LHS is:
[tex]\[ \text{LHS} = \frac{\sin^2(\beta) - \sin^2(a)}{\cos^2(\alpha) \cos^2(\beta)} \][/tex]
Next, let’s consider the right-hand side (RHS) of the equation:
### Step 3: Expand the RHS
Given:
[tex]\[ \text{RHS} = \tan^2(\beta) - \tan^2(\alpha) \][/tex]
We know that:
[tex]\[ \tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)} \][/tex]
Substituting this identity into the RHS, we have:
[tex]\[ \tan^2(\beta) - \tan^2(\alpha) = \frac{\sin^2(\beta)}{\cos^2(\beta)} - \frac{\sin^2(\alpha)}{\cos^2(\alpha)} \][/tex]
### Step 4: Compare the LHS and RHS
Now we need to check whether the transformed LHS matches the transformed RHS.
- LHS in its simplified form:
[tex]\[ \frac{\sin^2(\beta) - \sin^2(a)}{\cos^2(\alpha) \cos^2(\beta)} \][/tex]
- RHS in its transformed form:
[tex]\[ \frac{\sin^2(\beta)}{\cos^2(\beta)} - \frac{\sin^2(\alpha)}{\cos^2(\alpha)} \][/tex]
Simplifying the RHS further:
[tex]\[ \frac{\sin^2(\beta)}{\cos^2(\beta)} - \frac{\sin^2(\alpha)}{\cos^2(\alpha)} = \frac{\sin^2(\beta) \cos^2(\alpha) - \sin^2(\alpha) \cos^2(\beta)}{\cos^2(\alpha) \cos^2(\beta)} \][/tex]
Arriving at the same form as the LHS, thus verifying the identity.
Thus, we conclude:
[tex]\[ \frac{\cos^2(a) - \cos^2(\beta)}{\cos^2(\alpha) \cos^2(\beta)} = \tan^2(\beta) - \tan^2(\alpha) \][/tex]
