Question

match the expression on the left with an equivalent expression on the right. write the correct letter in the blank.
d 1.) $(1 + \cot(x))^2$ \t\t\t\t\ta.) $\csc(x)$
2.) $\frac{\cos(x)\tan(x)}{\sin(x)}$ \t\t\tb.) $(\sin(x) + \cot(x))(\sin(x) - \cot(x))$
3.) $\frac{-\sin^2(x) + 1}{-\cos(x)}$ \t\t\tc.) $\tan(x)(\cos(x) - 3)(\cos(x) + 1)$
4.) $\sin^2(x) - \cot^2(x)$ \t\t\t\td.) $\csc^2(x) + 2\cot(x)$
5.) $\cos(x) - \sec(x)$ \t\t\t\t\te.) $(2\sin(x) + 1)(\sin(x) - 3)$
6.) $2\sin^2(x) - 5\sin(x) - 3$ \t\t\tf.) $ - \sin(x)\tan(x)$
7.) $\cos^2(x)\tan(x) - 2\cos(x)\tan(x) - 3\tan(x)$ \tg.) $1$
8.) $\frac{\sec(x)}{\tan(x)}$ \t\t\t\t\th.) $ - \cos(x)$

Explanation:

Step1: Expand $(1+\cot x)^2$

$(1+\cot x)^2 = 1 + 2\cot x + \cot^2 x$. Since $1+\cot^2 x = \csc^2 x$, this becomes $\csc^2 x + 2\cot x$ (matches D).

Step2: Simplify $\frac{\cos x \tan x}{\sin x}$

$\tan x = \frac{\sin x}{\cos x}$, so $\cos x \cdot \frac{\sin x}{\cos x} = \sin x$. Then $\frac{\sin x}{\sin x} = 1$ (matches G).

Step3: Simplify $\frac{-\sin^2 x +1}{-\cos x}$

$1-\sin^2 x = \cos^2 x$, so $\frac{\cos^2 x}{-\cos x} = -\cos x$ (matches H).

Step4: Rewrite $\sin^2 x - \cot^2 x$

Difference of squares: $a^2 - b^2 = (a+b)(a-b)$, so $(\sin x + \cot x)(\sin x - \cot x)$ (matches B).

Step5: Simplify $\cos x - \sec x$

$\sec x = \frac{1}{\cos x}$, so $\cos x - \frac{1}{\cos x} = \frac{\cos^2 x -1}{\cos x} = \frac{-\sin^2 x}{\cos x} = -\sin x \cdot \frac{\sin x}{\cos x} = -\sin x \tan x$ (matches F).

Step6: Factor $2\sin^2 x -5\sin x -3$

Let $y=\sin x$: $2y^2 -5y -3=(2y+1)(y-3)$, so $(2\sin x +1)(\sin x -3)$ (matches E).

Step7: Factor $\cos^2 x \tan x -2\cos x \tan x -3\tan x$

Factor out $\tan x$: $\tan x(\cos^2 x -2\cos x -3) = \tan x(\cos x -3)(\cos x +1)$ (matches C).

Step8: Simplify $\frac{\sec x}{\tan x}$

$\sec x = \frac{1}{\cos x}$, $\tan x = \frac{\sin x}{\cos x}$, so $\frac{1/\cos x}{\sin x/\cos x} = \frac{1}{\sin x} = \csc x$ (matches A).

Answer:

1. D
2. G
3. H
4. B
5. F
6. E
7. C
8. A