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Sagot :
Let's analyze each expression to determine whether it is an identity.
### Expression A: [tex]\(\sin^2 x \cot^2 x + \cos^2 x \tan^2 x = 1\)[/tex]
1. Rewrite [tex]\(\cot x\)[/tex] and [tex]\(\tan x\)[/tex] in terms of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex]:
[tex]\[ \cot x = \frac{\cos x}{\sin x}, \quad \tan x = \frac{\sin x}{\cos x} \][/tex]
2. Substitute back into the equation:
[tex]\[ \sin^2 x \left(\frac{\cos x}{\sin x}\right)^2 + \cos^2 x \left(\frac{\sin x}{\cos x}\right)^2 \][/tex]
Simplify each term:
[tex]\[ \sin^2 x \cdot \frac{\cos^2 x}{\sin^2 x} + \cos^2 x \cdot \frac{\sin^2 x}{\cos^2 x} \][/tex]
[tex]\[ \cos^2 x + \sin^2 x \][/tex]
3. Knowing the Pythagorean identity, [tex]\(\cos^2 x + \sin^2 x = 1\)[/tex], this becomes:
[tex]\[ 1 = 1 \][/tex]
Thus, Expression A is an identity.
### Expression B: [tex]\(\tan^2 x + \cot^2 x = 1\)[/tex]
1. Rewrite [tex]\(\tan x\)[/tex] and [tex]\(\cot x\)[/tex] in terms of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex]:
[tex]\[ \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} \][/tex]
2. Substitute back into the equation:
[tex]\[ \left(\frac{\sin x}{\cos x}\right)^2 + \left(\frac{\cos x}{\sin x}\right)^2 \][/tex]
Simplify:
[tex]\[ \frac{\sin^2 x}{\cos^2 x} + \frac{\cos^2 x}{\sin^2 x} \][/tex]
3. The expression simplifies to:
[tex]\[ \frac{\sin^4 x + \cos^4 x}{\sin^2 x \cos^2 x} \][/tex]
Since [tex]\(\sin^4 x + \cos^4 x\)[/tex] does not simplify to [tex]\(\cos^2 x \sin^2 x\)[/tex], this expression is not identically equal to 1. Hence, Expression B is not an identity.
### Expression C: [tex]\((1 - 2 \sin^2 x) \csc^2 x = 4 \cos^2 x - 2\)[/tex]
1. Rewrite [tex]\(\csc x\)[/tex]:
[tex]\[ \csc x = \frac{1}{\sin x} \][/tex]
2. Substitute back into the equation:
[tex]\[ (1 - 2 \sin^2 x) \left(\frac{1}{\sin^2 x}\right) \][/tex]
3. Simplify:
[tex]\[ \frac{1 - 2 \sin^2 x}{\sin^2 x} = \frac{1}{\sin^2 x} - 2 \][/tex]
4. Rewrite [tex]\(\frac{1}{\sin^2 x}\)[/tex]:
[tex]\[ \csc^2 x = \frac{4 \cos^2 x - 2}{1} \][/tex]
Since [tex]\(\csc^2 x \left(1 - 2 \sin^2 x\right)\)[/tex] does not lead us to [tex]\(4 \cos^2 x - 2\)[/tex], this expression is not uniformly equivalent. Thus, Expression C is not an identity.
### Expression D: [tex]\(\sin x - \cos x + 1 = \tan x\)[/tex]
1. To check if this equation is valid, let’s examine the sides:
[tex]\[ \sin x - \cos x + 1 \][/tex]
2. Verify with special [tex]\(x\)[/tex] values:
- Let [tex]\(x = 0\)[/tex], then:
[tex]\[ \sin(0) - \cos(0) + 1 = 0 - 1 + 1 = 0 \neq \tan(0) = 0 \][/tex]
3. Use identities:
[tex]\[ \tan x = \frac{\sin x}{\cos x} \][/tex]
We clearly observe [tex]\(\sin x - \cos x + 1\)[/tex] does not directly simplify to [tex]\(\tan x\)[/tex]. Thus, Expression D is not an identity.
### Conclusion
Only Expression A, [tex]\(\sin^2 x \cot^2 x + \cos^2 x \tan^2 x = 1\)[/tex], is an identity.
Therefore, the correct answer is:
- [tex]\(\boxed{A}\)[/tex]
### Expression A: [tex]\(\sin^2 x \cot^2 x + \cos^2 x \tan^2 x = 1\)[/tex]
1. Rewrite [tex]\(\cot x\)[/tex] and [tex]\(\tan x\)[/tex] in terms of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex]:
[tex]\[ \cot x = \frac{\cos x}{\sin x}, \quad \tan x = \frac{\sin x}{\cos x} \][/tex]
2. Substitute back into the equation:
[tex]\[ \sin^2 x \left(\frac{\cos x}{\sin x}\right)^2 + \cos^2 x \left(\frac{\sin x}{\cos x}\right)^2 \][/tex]
Simplify each term:
[tex]\[ \sin^2 x \cdot \frac{\cos^2 x}{\sin^2 x} + \cos^2 x \cdot \frac{\sin^2 x}{\cos^2 x} \][/tex]
[tex]\[ \cos^2 x + \sin^2 x \][/tex]
3. Knowing the Pythagorean identity, [tex]\(\cos^2 x + \sin^2 x = 1\)[/tex], this becomes:
[tex]\[ 1 = 1 \][/tex]
Thus, Expression A is an identity.
### Expression B: [tex]\(\tan^2 x + \cot^2 x = 1\)[/tex]
1. Rewrite [tex]\(\tan x\)[/tex] and [tex]\(\cot x\)[/tex] in terms of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex]:
[tex]\[ \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} \][/tex]
2. Substitute back into the equation:
[tex]\[ \left(\frac{\sin x}{\cos x}\right)^2 + \left(\frac{\cos x}{\sin x}\right)^2 \][/tex]
Simplify:
[tex]\[ \frac{\sin^2 x}{\cos^2 x} + \frac{\cos^2 x}{\sin^2 x} \][/tex]
3. The expression simplifies to:
[tex]\[ \frac{\sin^4 x + \cos^4 x}{\sin^2 x \cos^2 x} \][/tex]
Since [tex]\(\sin^4 x + \cos^4 x\)[/tex] does not simplify to [tex]\(\cos^2 x \sin^2 x\)[/tex], this expression is not identically equal to 1. Hence, Expression B is not an identity.
### Expression C: [tex]\((1 - 2 \sin^2 x) \csc^2 x = 4 \cos^2 x - 2\)[/tex]
1. Rewrite [tex]\(\csc x\)[/tex]:
[tex]\[ \csc x = \frac{1}{\sin x} \][/tex]
2. Substitute back into the equation:
[tex]\[ (1 - 2 \sin^2 x) \left(\frac{1}{\sin^2 x}\right) \][/tex]
3. Simplify:
[tex]\[ \frac{1 - 2 \sin^2 x}{\sin^2 x} = \frac{1}{\sin^2 x} - 2 \][/tex]
4. Rewrite [tex]\(\frac{1}{\sin^2 x}\)[/tex]:
[tex]\[ \csc^2 x = \frac{4 \cos^2 x - 2}{1} \][/tex]
Since [tex]\(\csc^2 x \left(1 - 2 \sin^2 x\right)\)[/tex] does not lead us to [tex]\(4 \cos^2 x - 2\)[/tex], this expression is not uniformly equivalent. Thus, Expression C is not an identity.
### Expression D: [tex]\(\sin x - \cos x + 1 = \tan x\)[/tex]
1. To check if this equation is valid, let’s examine the sides:
[tex]\[ \sin x - \cos x + 1 \][/tex]
2. Verify with special [tex]\(x\)[/tex] values:
- Let [tex]\(x = 0\)[/tex], then:
[tex]\[ \sin(0) - \cos(0) + 1 = 0 - 1 + 1 = 0 \neq \tan(0) = 0 \][/tex]
3. Use identities:
[tex]\[ \tan x = \frac{\sin x}{\cos x} \][/tex]
We clearly observe [tex]\(\sin x - \cos x + 1\)[/tex] does not directly simplify to [tex]\(\tan x\)[/tex]. Thus, Expression D is not an identity.
### Conclusion
Only Expression A, [tex]\(\sin^2 x \cot^2 x + \cos^2 x \tan^2 x = 1\)[/tex], is an identity.
Therefore, the correct answer is:
- [tex]\(\boxed{A}\)[/tex]
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