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Simplify and solve the following equation:
[tex]\[ \frac{\cos^2 a - \cos^2 \beta}{\cos^2 \alpha \cdot \cos^2 \beta} = \tan^2 \beta - \tan^2 \alpha \][/tex]

Sagot :

GinnyAnswer
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]
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