Answer :
To determine which of the given statements is an identity, we need to check each option to see if it holds true for all values of [tex]\( x \)[/tex].
### Option A: [tex]\((\tan x + \cot x)(\sin x \cos x) = 1\)[/tex]
First, we express tan and cot in terms of sin and cos:
[tex]\[ \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} \][/tex]
Thus, the left-hand side can be rewritten as:
[tex]\[ \left( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right)(\sin x \cos x) \][/tex]
Combining the fractions:
[tex]\[ \left( \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \right)(\sin x \cos x) \][/tex]
Since [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex]:
[tex]\[ \frac{1}{\sin x \cos x} (\sin x \cos x) = 1 \][/tex]
The equation simplifies to:
[tex]\[ 1 = 1 \][/tex]
Thus, Option A is an identity.
### Option B: [tex]\(\sec x \tan x - \cos x \cot x = \sin x\)[/tex]
Express each trigonometric function in terms of sin and cos:
[tex]\[ \sec x = \frac{1}{\cos x}, \quad \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} \][/tex]
Then, the expression becomes:
[tex]\[ \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} - \cos x \cdot \frac{\cos x}{\sin x} \][/tex]
Simplify the product:
[tex]\[ \frac{\sin x}{\cos^2 x} - \frac{\cos^2 x}{\sin x} \][/tex]
Combine into a single fraction:
[tex]\[ \frac{\sin^2 x - \cos^4 x}{\cos^2 x \sin x} \][/tex]
This expression needs to simplify down to [tex]\(\sin x\)[/tex]:
[tex]\[ \frac{\sin^2 x - \cos^4 x}{\cos^2 x \sin x} \neq \sin x \][/tex]
Thus, Option B is not an identity.
### Option C: [tex]\(\cos x (2 \sin x + 1) = 0\)[/tex]
The expression involves a product of two terms. For the product to always be zero:
[tex]\[ \cos x = 0 \quad \text{or} \quad 2 \sin x + 1 = 0 \][/tex]
[tex]\(\cos x = 0\)[/tex] happens at [tex]\(x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots\)[/tex]
[tex]\[ 2 \sin x + 1 = 0 \rightarrow \sin x = -\frac{1}{2} \][/tex]
[tex]\(\sin x = -\frac{1}{2}\)[/tex] happens at [tex]\(x = \frac{7\pi}{6}, \frac{11\pi}{6}, \ldots\)[/tex]
Since not every value of [tex]\(x\)[/tex] will satisfy either equation, it is not an identity.
### Option D: [tex]\(\sin^2 x = 4 - 2 \cos^2 x\)[/tex]
We know from the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \implies \sin^2 x = 1 - \cos^2 x \][/tex]
Check the given equation:
[tex]\[ 1 - \cos^2 x = 4 - 2 \cos^2 x \][/tex]
Rearrange to see if it is true:
[tex]\[ 1 - \cos^2 x = 4 - 2 \cos^2 x \implies \cos^2 x = 3 \][/tex]
This is not possible since [tex]\(\cos^2 x\)[/tex] must be between 0 and 1. Thus, Option D is not an identity.
### Conclusion: The only equation that holds true for all values of [tex]\(x\)[/tex] is
[tex]\[ \boxed{A} \][/tex]
### Option A: [tex]\((\tan x + \cot x)(\sin x \cos x) = 1\)[/tex]
First, we express tan and cot in terms of sin and cos:
[tex]\[ \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} \][/tex]
Thus, the left-hand side can be rewritten as:
[tex]\[ \left( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right)(\sin x \cos x) \][/tex]
Combining the fractions:
[tex]\[ \left( \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \right)(\sin x \cos x) \][/tex]
Since [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex]:
[tex]\[ \frac{1}{\sin x \cos x} (\sin x \cos x) = 1 \][/tex]
The equation simplifies to:
[tex]\[ 1 = 1 \][/tex]
Thus, Option A is an identity.
### Option B: [tex]\(\sec x \tan x - \cos x \cot x = \sin x\)[/tex]
Express each trigonometric function in terms of sin and cos:
[tex]\[ \sec x = \frac{1}{\cos x}, \quad \tan x = \frac{\sin x}{\cos x}, \quad \cot x = \frac{\cos x}{\sin x} \][/tex]
Then, the expression becomes:
[tex]\[ \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} - \cos x \cdot \frac{\cos x}{\sin x} \][/tex]
Simplify the product:
[tex]\[ \frac{\sin x}{\cos^2 x} - \frac{\cos^2 x}{\sin x} \][/tex]
Combine into a single fraction:
[tex]\[ \frac{\sin^2 x - \cos^4 x}{\cos^2 x \sin x} \][/tex]
This expression needs to simplify down to [tex]\(\sin x\)[/tex]:
[tex]\[ \frac{\sin^2 x - \cos^4 x}{\cos^2 x \sin x} \neq \sin x \][/tex]
Thus, Option B is not an identity.
### Option C: [tex]\(\cos x (2 \sin x + 1) = 0\)[/tex]
The expression involves a product of two terms. For the product to always be zero:
[tex]\[ \cos x = 0 \quad \text{or} \quad 2 \sin x + 1 = 0 \][/tex]
[tex]\(\cos x = 0\)[/tex] happens at [tex]\(x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots\)[/tex]
[tex]\[ 2 \sin x + 1 = 0 \rightarrow \sin x = -\frac{1}{2} \][/tex]
[tex]\(\sin x = -\frac{1}{2}\)[/tex] happens at [tex]\(x = \frac{7\pi}{6}, \frac{11\pi}{6}, \ldots\)[/tex]
Since not every value of [tex]\(x\)[/tex] will satisfy either equation, it is not an identity.
### Option D: [tex]\(\sin^2 x = 4 - 2 \cos^2 x\)[/tex]
We know from the Pythagorean identity:
[tex]\[ \sin^2 x + \cos^2 x = 1 \implies \sin^2 x = 1 - \cos^2 x \][/tex]
Check the given equation:
[tex]\[ 1 - \cos^2 x = 4 - 2 \cos^2 x \][/tex]
Rearrange to see if it is true:
[tex]\[ 1 - \cos^2 x = 4 - 2 \cos^2 x \implies \cos^2 x = 3 \][/tex]
This is not possible since [tex]\(\cos^2 x\)[/tex] must be between 0 and 1. Thus, Option D is not an identity.
### Conclusion: The only equation that holds true for all values of [tex]\(x\)[/tex] is
[tex]\[ \boxed{A} \][/tex]
