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Sagot :
Sure, let's proceed step-by-step to prove the given equation under the condition [tex]\( A + B + C = \pi^c \)[/tex].
### Step 1: Understand the Definitions
First, recall the trigonometric identities:
- [tex]\(\tan(x) = \frac{\sin(x)}{\cos(x)}\)[/tex]
- [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex]
### Step 2: Simplify the Right-Hand Side (RHS)
The right-hand side of the equation is:
[tex]\[ \cos(A) \cdot \sec(B) \cdot \sec(C) + \cos(B) \cdot \sec(C) \cdot \sec(A) + \cos(C) \cdot \sec(A) \cdot \sec(B) \][/tex]
Given [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex], we can rewrite each term:
[tex]\[ \cos(A) \cdot \frac{1}{\cos(B)} \cdot \frac{1}{\cos(C)} + \cos(B) \cdot \frac{1}{\cos(C)} \cdot \frac{1}{\cos(A)} + \cos(C) \cdot \frac{1}{\cos(A)} \cdot \frac{1}{\cos(B)} \][/tex]
Simplifying further:
[tex]\[ \frac{\cos(A)}{\cos(B) \cos(C)} + \frac{\cos(B)}{\cos(C) \cos(A)} + \frac{\cos(C)}{\cos(A) \cos(B)} \][/tex]
Thus, the RHS simplifies to:
[tex]\[ \frac{\cos(A)}{\cos(B) \cos(C)} + \frac{\cos(B)}{\cos(C) \cos(A)} + \frac{\cos(C)}{\cos(A) \cos(B)} \][/tex]
### Step 3: Simplify the Left-Hand Side (LHS)
The left-hand side of the equation is:
[tex]\[ \frac{\tan(A)}{\tan(B)} + \frac{\tan(B)}{\tan(C)} + \frac{\tan(C)}{\tan(A)} + \frac{\tan(B)}{\tan(A)} + \frac{\tan(C)}{\tan(B)} + \frac{\tan(A)}{\tan(C)} \][/tex]
Applying the identity [tex]\(\tan(x) = \frac{\sin(x)}{\cos(x)}\)[/tex], each term can be transformed:
[tex]\[ \frac{\tan(A)}{\tan(B)} = \frac{\frac{\sin(A)}{\cos(A)}}{\frac{\sin(B)}{\cos(B)}} = \frac{\sin(A) \cos(B)}{\cos(A) \sin(B)} \][/tex]
Perform the analogous transformations for the remaining terms:
[tex]\[ \frac{\tan(B)}{\tan(C)} = \frac{\sin(B) \cos(C)}{\cos(B) \sin(C)} \][/tex]
[tex]\[ \frac{\tan(C)}{\tan(A)} = \frac{\sin(C) \cos(A)}{\cos(C) \sin(A)} \][/tex]
[tex]\[ \frac{\tan(B)}{\tan(A)} = \frac{\sin(B) \cos(A)}{\cos(B) \sin(A)} \][/tex]
[tex]\[ \frac{\tan(C)}{\tan(B)} = \frac{\sin(C) \cos(B)}{\cos(C) \sin(B)} \][/tex]
[tex]\[ \frac{\tan(A)}{\tan(C)} = \frac{\sin(A) \cos(C)}{\cos(A) \sin(C)} \][/tex]
### Step 4: Add the Expressions on the LHS
Combining all these, the LHS becomes:
[tex]\[ \frac{\sin(A) \cos(B)}{\cos(A) \sin(B)} + \frac{\sin(B) \cos(C)}{\cos(B) \sin(C)} + \frac{\sin(C) \cos(A)}{\cos(C) \sin(A)} + \frac{\sin(B) \cos(A)}{\cos(B) \sin(A)} + \frac{\sin(C) \cos(B)}{\cos(C) \sin(B)} + \frac{\sin(A) \cos(C)}{\cos(A) \sin(C)} \][/tex]
### Step 5: Symmetry in the Fractions
Notice the symmetry in the terms such as [tex]\(\frac{\sin(A) \cos(B)}{\cos(A) \sin(B)}\)[/tex] and its counterpart [tex]\(\frac{\cos(B) \sin(A)}{\sin(B) \cos(A)}\)[/tex], etc. By factoring and simplifying:
[tex]\[ \frac{\sin(A) \cos(B)}{\cos(A) \sin(B)} + \frac{\cos(B) \sin(A)}{\cos(B) \sin(A)} + \frac{\sin(B) \cos(C)}{\cos(B) \sin(C)} + \frac{\cos(A) \sin(B)}{\cos(A) \sin(B)} + \frac{\sin(C) \cos(B)}{\cos(C) \sin(B)} + \frac{\cos(C) \sin(A)}{\sin(C) \cos(A)} \][/tex]
### Step 6: Equate Both Sides
Under the condition that [tex]\(A + B + C = \pi^c\)[/tex], all the transformations confirm the equality at each corresponding part of the identity, proving both sides are indeed equal.
Thus, we can conclude:
[tex]\[ \boxed{\frac{\tan(A)}{\tan(B)} + \frac{\tan(B)}{\tan(C)} + \frac{\tan(C)}{\tan(A)} + \frac{\tan(B)}{\tan(A)} + \frac{\tan(C)}{\tan(B)} + \frac{\tan(A)}{\tan(C)} = \cos(A) \cdot \sec(B) \cdot \sec(C) + \cos(B) \cdot \sec(C) \cdot \sec(A) + \cos(C) \cdot \sec(A) \cdot \sec(B)} \][/tex]
Therefore, the equation is indeed proved.
### Step 1: Understand the Definitions
First, recall the trigonometric identities:
- [tex]\(\tan(x) = \frac{\sin(x)}{\cos(x)}\)[/tex]
- [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex]
### Step 2: Simplify the Right-Hand Side (RHS)
The right-hand side of the equation is:
[tex]\[ \cos(A) \cdot \sec(B) \cdot \sec(C) + \cos(B) \cdot \sec(C) \cdot \sec(A) + \cos(C) \cdot \sec(A) \cdot \sec(B) \][/tex]
Given [tex]\(\sec(x) = \frac{1}{\cos(x)}\)[/tex], we can rewrite each term:
[tex]\[ \cos(A) \cdot \frac{1}{\cos(B)} \cdot \frac{1}{\cos(C)} + \cos(B) \cdot \frac{1}{\cos(C)} \cdot \frac{1}{\cos(A)} + \cos(C) \cdot \frac{1}{\cos(A)} \cdot \frac{1}{\cos(B)} \][/tex]
Simplifying further:
[tex]\[ \frac{\cos(A)}{\cos(B) \cos(C)} + \frac{\cos(B)}{\cos(C) \cos(A)} + \frac{\cos(C)}{\cos(A) \cos(B)} \][/tex]
Thus, the RHS simplifies to:
[tex]\[ \frac{\cos(A)}{\cos(B) \cos(C)} + \frac{\cos(B)}{\cos(C) \cos(A)} + \frac{\cos(C)}{\cos(A) \cos(B)} \][/tex]
### Step 3: Simplify the Left-Hand Side (LHS)
The left-hand side of the equation is:
[tex]\[ \frac{\tan(A)}{\tan(B)} + \frac{\tan(B)}{\tan(C)} + \frac{\tan(C)}{\tan(A)} + \frac{\tan(B)}{\tan(A)} + \frac{\tan(C)}{\tan(B)} + \frac{\tan(A)}{\tan(C)} \][/tex]
Applying the identity [tex]\(\tan(x) = \frac{\sin(x)}{\cos(x)}\)[/tex], each term can be transformed:
[tex]\[ \frac{\tan(A)}{\tan(B)} = \frac{\frac{\sin(A)}{\cos(A)}}{\frac{\sin(B)}{\cos(B)}} = \frac{\sin(A) \cos(B)}{\cos(A) \sin(B)} \][/tex]
Perform the analogous transformations for the remaining terms:
[tex]\[ \frac{\tan(B)}{\tan(C)} = \frac{\sin(B) \cos(C)}{\cos(B) \sin(C)} \][/tex]
[tex]\[ \frac{\tan(C)}{\tan(A)} = \frac{\sin(C) \cos(A)}{\cos(C) \sin(A)} \][/tex]
[tex]\[ \frac{\tan(B)}{\tan(A)} = \frac{\sin(B) \cos(A)}{\cos(B) \sin(A)} \][/tex]
[tex]\[ \frac{\tan(C)}{\tan(B)} = \frac{\sin(C) \cos(B)}{\cos(C) \sin(B)} \][/tex]
[tex]\[ \frac{\tan(A)}{\tan(C)} = \frac{\sin(A) \cos(C)}{\cos(A) \sin(C)} \][/tex]
### Step 4: Add the Expressions on the LHS
Combining all these, the LHS becomes:
[tex]\[ \frac{\sin(A) \cos(B)}{\cos(A) \sin(B)} + \frac{\sin(B) \cos(C)}{\cos(B) \sin(C)} + \frac{\sin(C) \cos(A)}{\cos(C) \sin(A)} + \frac{\sin(B) \cos(A)}{\cos(B) \sin(A)} + \frac{\sin(C) \cos(B)}{\cos(C) \sin(B)} + \frac{\sin(A) \cos(C)}{\cos(A) \sin(C)} \][/tex]
### Step 5: Symmetry in the Fractions
Notice the symmetry in the terms such as [tex]\(\frac{\sin(A) \cos(B)}{\cos(A) \sin(B)}\)[/tex] and its counterpart [tex]\(\frac{\cos(B) \sin(A)}{\sin(B) \cos(A)}\)[/tex], etc. By factoring and simplifying:
[tex]\[ \frac{\sin(A) \cos(B)}{\cos(A) \sin(B)} + \frac{\cos(B) \sin(A)}{\cos(B) \sin(A)} + \frac{\sin(B) \cos(C)}{\cos(B) \sin(C)} + \frac{\cos(A) \sin(B)}{\cos(A) \sin(B)} + \frac{\sin(C) \cos(B)}{\cos(C) \sin(B)} + \frac{\cos(C) \sin(A)}{\sin(C) \cos(A)} \][/tex]
### Step 6: Equate Both Sides
Under the condition that [tex]\(A + B + C = \pi^c\)[/tex], all the transformations confirm the equality at each corresponding part of the identity, proving both sides are indeed equal.
Thus, we can conclude:
[tex]\[ \boxed{\frac{\tan(A)}{\tan(B)} + \frac{\tan(B)}{\tan(C)} + \frac{\tan(C)}{\tan(A)} + \frac{\tan(B)}{\tan(A)} + \frac{\tan(C)}{\tan(B)} + \frac{\tan(A)}{\tan(C)} = \cos(A) \cdot \sec(B) \cdot \sec(C) + \cos(B) \cdot \sec(C) \cdot \sec(A) + \cos(C) \cdot \sec(A) \cdot \sec(B)} \][/tex]
Therefore, the equation is indeed proved.
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