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Sagot :
Let's evaluate each of the given trigonometric equations to determine which are true identities and which are not.
A. [tex]\(\frac{\sec x - \cos x}{\sec x} = \sin^2 x\)[/tex]
First, express [tex]\(\sec x\)[/tex] in terms of [tex]\(\cos x\)[/tex]:
[tex]\[ \sec x = \frac{1}{\cos x} \][/tex]
Rewrite the left-hand side:
[tex]\[ \frac{\sec x - \cos x}{\sec x} = \frac{\frac{1}{\cos x} - \cos x}{\frac{1}{\cos x}} \][/tex]
Simplify:
[tex]\[ \frac{\frac{1 - \cos^2 x}{\cos x}}{\frac{1}{\cos x}} = \frac{1 - \cos^2 x}{\cos x} \cdot \cos x = 1 - \cos^2 x \][/tex]
Recall the Pythagorean identity [tex]\(1 - \cos^2 x = \sin^2 x\)[/tex]:
[tex]\[ 1 - \cos^2 x = \sin^2 x \][/tex]
So the left-hand side simplifies to:
[tex]\[ \sin^2 x = \sin^2 x \][/tex]
This confirms that equation A is a valid trigonometric identity.
B. [tex]\(\tan x \cos x \csc x = 1\)[/tex]
First, express [tex]\(\tan x\)[/tex] and [tex]\(\csc x\)[/tex] in terms of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex]:
[tex]\[ \tan x = \frac{\sin x}{\cos x}, \quad \csc x = \frac{1}{\sin x} \][/tex]
Rewrite the left-hand side:
[tex]\[ \tan x \cos x \csc x = \left(\frac{\sin x}{\cos x}\right) \cos x \left(\frac{1}{\sin x}\right) \][/tex]
Simplify:
[tex]\[ \frac{\sin x}{\cos x} \cdot \cos x \cdot \frac{1}{\sin x} = \frac{\sin x \cos x}{\cos x \sin x} = 1 \][/tex]
This confirms that equation B is a valid trigonometric identity.
C. [tex]\(4 \cos x \sin x = 2 \cos x + 1 - 2 \sin x\)[/tex]
Simplify each side separately.
The left-hand side is:
[tex]\[ 4 \cos x \sin x \][/tex]
The right-hand side is:
[tex]\[ 2 \cos x + 1 - 2 \sin x \][/tex]
Rewriting [tex]\(4 \cos x \sin x\)[/tex] using the double-angle identity ([tex]\(\sin(2x) = 2 \sin x \cos x\)[/tex]):
[tex]\[ 4 \cos x \sin x = 2 (2 \cos x \sin x) = 2 \sin(2x) \][/tex]
Substituting into the equation, we get:
[tex]\[ 2 \sin(2x) = 2 \cos x + 1 - 2 \sin x \][/tex]
There is no known trigonometric identity that matches [tex]\(2 \sin(2x)\)[/tex] to [tex]\(2 \cos x + 1 - 2 \sin x\)[/tex]:
This proves that equation C is not a valid trigonometric identity.
D. [tex]\(1 - \tan x \tan y = \frac{\cos (x + y)}{\cos x \cos y}\)[/tex]
Using the sum formula for cosine:
[tex]\[ \cos(x + y) = \cos x \cos y - \sin x \sin y \][/tex]
Rewrite the right-hand side:
[tex]\[ \frac{\cos(x + y)}{\cos x \cos y} = \frac{\cos x \cos y - \sin x \sin y}{\cos x \cos y} \][/tex]
Simplify:
[tex]\[ = \frac{\cos x \cos y}{\cos x \cos y} - \frac{\sin x \sin y}{\cos x \cos y} = 1 - \tan x \tan y \][/tex]
This confirms that equation D is a valid trigonometric identity.
Summary:
- A: [tex]\(\frac{\sec x - \cos x}{\sec x} = \sin^2 x\)[/tex] is a trigonometric identity.
- B: [tex]\(\tan x \cos x \csc x = 1\)[/tex] is a trigonometric identity.
- C: [tex]\(4 \cos x \sin x = 2 \cos x + 1 - 2 \sin x\)[/tex] is not a trigonometric identity.
- D: [tex]\(1 - \tan x \tan y = \frac{\cos (x + y)}{\cos x \cos y}\)[/tex] is a trigonometric identity.
Thus, the equation which is not a trigonometric identity is:
- C.
A. [tex]\(\frac{\sec x - \cos x}{\sec x} = \sin^2 x\)[/tex]
First, express [tex]\(\sec x\)[/tex] in terms of [tex]\(\cos x\)[/tex]:
[tex]\[ \sec x = \frac{1}{\cos x} \][/tex]
Rewrite the left-hand side:
[tex]\[ \frac{\sec x - \cos x}{\sec x} = \frac{\frac{1}{\cos x} - \cos x}{\frac{1}{\cos x}} \][/tex]
Simplify:
[tex]\[ \frac{\frac{1 - \cos^2 x}{\cos x}}{\frac{1}{\cos x}} = \frac{1 - \cos^2 x}{\cos x} \cdot \cos x = 1 - \cos^2 x \][/tex]
Recall the Pythagorean identity [tex]\(1 - \cos^2 x = \sin^2 x\)[/tex]:
[tex]\[ 1 - \cos^2 x = \sin^2 x \][/tex]
So the left-hand side simplifies to:
[tex]\[ \sin^2 x = \sin^2 x \][/tex]
This confirms that equation A is a valid trigonometric identity.
B. [tex]\(\tan x \cos x \csc x = 1\)[/tex]
First, express [tex]\(\tan x\)[/tex] and [tex]\(\csc x\)[/tex] in terms of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex]:
[tex]\[ \tan x = \frac{\sin x}{\cos x}, \quad \csc x = \frac{1}{\sin x} \][/tex]
Rewrite the left-hand side:
[tex]\[ \tan x \cos x \csc x = \left(\frac{\sin x}{\cos x}\right) \cos x \left(\frac{1}{\sin x}\right) \][/tex]
Simplify:
[tex]\[ \frac{\sin x}{\cos x} \cdot \cos x \cdot \frac{1}{\sin x} = \frac{\sin x \cos x}{\cos x \sin x} = 1 \][/tex]
This confirms that equation B is a valid trigonometric identity.
C. [tex]\(4 \cos x \sin x = 2 \cos x + 1 - 2 \sin x\)[/tex]
Simplify each side separately.
The left-hand side is:
[tex]\[ 4 \cos x \sin x \][/tex]
The right-hand side is:
[tex]\[ 2 \cos x + 1 - 2 \sin x \][/tex]
Rewriting [tex]\(4 \cos x \sin x\)[/tex] using the double-angle identity ([tex]\(\sin(2x) = 2 \sin x \cos x\)[/tex]):
[tex]\[ 4 \cos x \sin x = 2 (2 \cos x \sin x) = 2 \sin(2x) \][/tex]
Substituting into the equation, we get:
[tex]\[ 2 \sin(2x) = 2 \cos x + 1 - 2 \sin x \][/tex]
There is no known trigonometric identity that matches [tex]\(2 \sin(2x)\)[/tex] to [tex]\(2 \cos x + 1 - 2 \sin x\)[/tex]:
This proves that equation C is not a valid trigonometric identity.
D. [tex]\(1 - \tan x \tan y = \frac{\cos (x + y)}{\cos x \cos y}\)[/tex]
Using the sum formula for cosine:
[tex]\[ \cos(x + y) = \cos x \cos y - \sin x \sin y \][/tex]
Rewrite the right-hand side:
[tex]\[ \frac{\cos(x + y)}{\cos x \cos y} = \frac{\cos x \cos y - \sin x \sin y}{\cos x \cos y} \][/tex]
Simplify:
[tex]\[ = \frac{\cos x \cos y}{\cos x \cos y} - \frac{\sin x \sin y}{\cos x \cos y} = 1 - \tan x \tan y \][/tex]
This confirms that equation D is a valid trigonometric identity.
Summary:
- A: [tex]\(\frac{\sec x - \cos x}{\sec x} = \sin^2 x\)[/tex] is a trigonometric identity.
- B: [tex]\(\tan x \cos x \csc x = 1\)[/tex] is a trigonometric identity.
- C: [tex]\(4 \cos x \sin x = 2 \cos x + 1 - 2 \sin x\)[/tex] is not a trigonometric identity.
- D: [tex]\(1 - \tan x \tan y = \frac{\cos (x + y)}{\cos x \cos y}\)[/tex] is a trigonometric identity.
Thus, the equation which is not a trigonometric identity is:
- C.
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