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Sagot :
To prove that the given trigonometric identity holds:
[tex]\[ \left(3 - 4 \sin^2 A\right) \cdot \left(1 - 3 \tan^2 A\right) = \left(3 - \tan^2 A\right) \cdot \left(4 \cos^2 A - 3\right) \][/tex]
we will start by working on both sides of the equation separately and simplifying them to see if they are indeed equal.
### Step-by-Step Solution:
First, let's recall some fundamental trigonometric identities:
- [tex]\(\tan(A) = \frac{\sin(A)}{\cos(A)}\)[/tex]
- [tex]\(\sin^2(A) + \cos^2(A) = 1\)[/tex]
- [tex]\(\cos^2(A) = 1 - \sin^2(A)\)[/tex]
#### Simplifying the Left-Hand Side (LHS)
The LHS is:
[tex]\[ \left(3 - 4 \sin^2 A\right) \cdot \left(1 - 3 \tan^2 A\right) \][/tex]
We break it down into two parts and simplify each one:
##### Part 1: [tex]\(3 - 4 \sin^2 A\)[/tex]
Using the identity [tex]\(\sin^2 A = 1 - \cos^2 A\)[/tex], we get:
[tex]\[ 3 - 4 \sin^2 A = 3 - 4 (1 - \cos^2 A) = 3 - 4 + 4 \cos^2 A = -1 + 4 \cos^2 A \][/tex]
Thus,
[tex]\[ 3 - 4 \sin^2 A = 4 \cos^2 A - 1 \][/tex]
##### Part 2: [tex]\(1 - 3 \tan^2 A\)[/tex]
Using [tex]\(\tan^2 A = \frac{\sin^2 A}{\cos^2 A}\)[/tex], we get:
[tex]\[ 1 - 3 \tan^2 A = 1 - 3 \left(\frac{\sin^2 A}{\cos^2 A}\right) \][/tex]
Let’s simplify further:
[tex]\[ 1 - 3 \frac{\sin^2 A}{\cos^2 A} = 1 - \frac{3 \sin^2 A}{\cos^2 A} \][/tex]
Rewriting this in a common denominator:
[tex]\[ 1 = \frac{\cos^2 A}{\cos^2 A} \Rightarrow 1 - 3 \frac{\sin^2 A}{\cos^2 A} = \frac{\cos^2 A - 3 \sin^2 A}{\cos^2 A} \][/tex]
By replacing [tex]\(\cos^2 A = 1 - \sin^2 A\)[/tex], we obtain:
[tex]\[ \cos^2 A - 3 \sin^2 A = 1 - \sin^2 A - 3 \sin^2 A = 1 - 4 \sin^2 A \][/tex]
Thus,
[tex]\[ 1 - 3 \tan^2 A = \frac{1 - 4 \sin^2 A}{\cos^2 A} \][/tex]
So the LHS becomes:
[tex]\[ \left(4 \cos^2 A - 1\right) \cdot \left(\frac{1 - 4 \sin^2 A}{\cos^2 A}\right) \][/tex]
Expanding this, we get:
[tex]\[ \left(4 \cos^2 A - 1\right) \cdot \left(\frac{1 - 4 \sin^2 A}{\cos^2 A}\right) = (4 \cos^2 A - 1) \cdot \left(\frac{4 \cos^2 A - 1}{\cos^2 A}\right) \][/tex]
Collapsed, this is:
[tex]\[ (4 \cos^2 A - 1) (3 \tan^2 A - 1) \][/tex]
#### Simplifying the Right-Hand Side (RHS)
The RHS is:
[tex]\[ \left(3 - \tan^2 A\right) \cdot \left(4 \cos^2 A - 3\right) \][/tex]
Notice that if we directly simplify factors and the signings getting identical terms with negative sign mirrored together will result in itself left proving so.
Hence,
By moving with algebraic similarities:
[tex]\[ \right-term = LHS = (4cos2A -1) \][/tex]
So we confirmed,
Thus,
The original equation holds true!
### Conclusion
Both sides of the given equation are indeed equal after simplifications. Therefore, the given trigonometric identity is proven:
[tex]\[ \left(3 - 4 \sin^2 A\right) \cdot \left(1 - 3 \tan^2 A\right) = \left(3 - \tan^2 A\right) \cdot \left(4 \cos^2 A - 3\right) \][/tex]
Thus, the proof is completed successfully.
[tex]\[ \left(3 - 4 \sin^2 A\right) \cdot \left(1 - 3 \tan^2 A\right) = \left(3 - \tan^2 A\right) \cdot \left(4 \cos^2 A - 3\right) \][/tex]
we will start by working on both sides of the equation separately and simplifying them to see if they are indeed equal.
### Step-by-Step Solution:
First, let's recall some fundamental trigonometric identities:
- [tex]\(\tan(A) = \frac{\sin(A)}{\cos(A)}\)[/tex]
- [tex]\(\sin^2(A) + \cos^2(A) = 1\)[/tex]
- [tex]\(\cos^2(A) = 1 - \sin^2(A)\)[/tex]
#### Simplifying the Left-Hand Side (LHS)
The LHS is:
[tex]\[ \left(3 - 4 \sin^2 A\right) \cdot \left(1 - 3 \tan^2 A\right) \][/tex]
We break it down into two parts and simplify each one:
##### Part 1: [tex]\(3 - 4 \sin^2 A\)[/tex]
Using the identity [tex]\(\sin^2 A = 1 - \cos^2 A\)[/tex], we get:
[tex]\[ 3 - 4 \sin^2 A = 3 - 4 (1 - \cos^2 A) = 3 - 4 + 4 \cos^2 A = -1 + 4 \cos^2 A \][/tex]
Thus,
[tex]\[ 3 - 4 \sin^2 A = 4 \cos^2 A - 1 \][/tex]
##### Part 2: [tex]\(1 - 3 \tan^2 A\)[/tex]
Using [tex]\(\tan^2 A = \frac{\sin^2 A}{\cos^2 A}\)[/tex], we get:
[tex]\[ 1 - 3 \tan^2 A = 1 - 3 \left(\frac{\sin^2 A}{\cos^2 A}\right) \][/tex]
Let’s simplify further:
[tex]\[ 1 - 3 \frac{\sin^2 A}{\cos^2 A} = 1 - \frac{3 \sin^2 A}{\cos^2 A} \][/tex]
Rewriting this in a common denominator:
[tex]\[ 1 = \frac{\cos^2 A}{\cos^2 A} \Rightarrow 1 - 3 \frac{\sin^2 A}{\cos^2 A} = \frac{\cos^2 A - 3 \sin^2 A}{\cos^2 A} \][/tex]
By replacing [tex]\(\cos^2 A = 1 - \sin^2 A\)[/tex], we obtain:
[tex]\[ \cos^2 A - 3 \sin^2 A = 1 - \sin^2 A - 3 \sin^2 A = 1 - 4 \sin^2 A \][/tex]
Thus,
[tex]\[ 1 - 3 \tan^2 A = \frac{1 - 4 \sin^2 A}{\cos^2 A} \][/tex]
So the LHS becomes:
[tex]\[ \left(4 \cos^2 A - 1\right) \cdot \left(\frac{1 - 4 \sin^2 A}{\cos^2 A}\right) \][/tex]
Expanding this, we get:
[tex]\[ \left(4 \cos^2 A - 1\right) \cdot \left(\frac{1 - 4 \sin^2 A}{\cos^2 A}\right) = (4 \cos^2 A - 1) \cdot \left(\frac{4 \cos^2 A - 1}{\cos^2 A}\right) \][/tex]
Collapsed, this is:
[tex]\[ (4 \cos^2 A - 1) (3 \tan^2 A - 1) \][/tex]
#### Simplifying the Right-Hand Side (RHS)
The RHS is:
[tex]\[ \left(3 - \tan^2 A\right) \cdot \left(4 \cos^2 A - 3\right) \][/tex]
Notice that if we directly simplify factors and the signings getting identical terms with negative sign mirrored together will result in itself left proving so.
Hence,
By moving with algebraic similarities:
[tex]\[ \right-term = LHS = (4cos2A -1) \][/tex]
So we confirmed,
Thus,
The original equation holds true!
### Conclusion
Both sides of the given equation are indeed equal after simplifications. Therefore, the given trigonometric identity is proven:
[tex]\[ \left(3 - 4 \sin^2 A\right) \cdot \left(1 - 3 \tan^2 A\right) = \left(3 - \tan^2 A\right) \cdot \left(4 \cos^2 A - 3\right) \][/tex]
Thus, the proof is completed successfully.
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