Using Sum and Product Identities - Math

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Question

Simplify $\sin x\cot x$ .

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Answer

To simplify $\sin x\cot x$ , break them into their SOHCAHTOA parts:

$$\frac{\text{opposite}$}{$\text{hypotenuse}$}*$\frac{\text{adjacent}$}{$\text{opposite}$}$ .

Notice that the opposite's cancel out, leaving $$\frac{\text{adjacent}$}{$\text{hypotenuse}$}=\cos x$ .

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Question

Simplify $\csc x \tan x$ .

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Answer

Break down $\csc x \tan x$ into SOHCAHTOA to solve:

$\csc=\frac{\text{hypotenuse}$}{$\text{opposite}$}$ and $\tan=\frac{\text{opposite}$}{$\text{adjacent}$}$ .

Therefore, $\csc x \tan x=\frac{\text{hypotenuse}$}{$\text{opposite}$}*$\frac{\text{opposite}$}{$\text{adjacent}$}$ . Note that the opposite's cancel out, leaving $$\frac{\text{hypotenuse}$}{$\text{adjacent}$}$ , which is the same as $\sec x$ .

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Question

Simplify $\sin x\cot x$ .

Tap to reveal answer

Answer

To simplify $\sin x\cot x$ , break them into their SOHCAHTOA parts:

$$\frac{\text{opposite}$}{$\text{hypotenuse}$}*$\frac{\text{adjacent}$}{$\text{opposite}$}$ .

Notice that the opposite's cancel out, leaving $$\frac{\text{adjacent}$}{$\text{hypotenuse}$}=\cos x$ .

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Question

Simplify $\csc x \tan x$ .

Tap to reveal answer

Answer

Break down $\csc x \tan x$ into SOHCAHTOA to solve:

$\csc=\frac{\text{hypotenuse}$}{$\text{opposite}$}$ and $\tan=\frac{\text{opposite}$}{$\text{adjacent}$}$ .

Therefore, $\csc x \tan x=\frac{\text{hypotenuse}$}{$\text{opposite}$}*$\frac{\text{opposite}$}{$\text{adjacent}$}$ . Note that the opposite's cancel out, leaving $$\frac{\text{hypotenuse}$}{$\text{adjacent}$}$ , which is the same as $\sec x$ .

← Didn't Know|Knew It →

Question

Simplify $\sin x\cot x$ .

Tap to reveal answer

Answer

To simplify $\sin x\cot x$ , break them into their SOHCAHTOA parts:

$$\frac{\text{opposite}$}{$\text{hypotenuse}$}*$\frac{\text{adjacent}$}{$\text{opposite}$}$ .

Notice that the opposite's cancel out, leaving $$\frac{\text{adjacent}$}{$\text{hypotenuse}$}=\cos x$ .

← Didn't Know|Knew It →

Question

Simplify $\csc x \tan x$ .

Tap to reveal answer

Answer

Break down $\csc x \tan x$ into SOHCAHTOA to solve:

$\csc=\frac{\text{hypotenuse}$}{$\text{opposite}$}$ and $\tan=\frac{\text{opposite}$}{$\text{adjacent}$}$ .

Therefore, $\csc x \tan x=\frac{\text{hypotenuse}$}{$\text{opposite}$}*$\frac{\text{opposite}$}{$\text{adjacent}$}$ . Note that the opposite's cancel out, leaving $$\frac{\text{hypotenuse}$}{$\text{adjacent}$}$ , which is the same as $\sec x$ .

← Didn't Know|Knew It →

Question

Simplify $\sin x\cot x$ .

Tap to reveal answer

Answer

To simplify $\sin x\cot x$ , break them into their SOHCAHTOA parts:

$$\frac{\text{opposite}$}{$\text{hypotenuse}$}*$\frac{\text{adjacent}$}{$\text{opposite}$}$ .

Notice that the opposite's cancel out, leaving $$\frac{\text{adjacent}$}{$\text{hypotenuse}$}=\cos x$ .

← Didn't Know|Knew It →

Question

Simplify $\csc x \tan x$ .

Tap to reveal answer

Answer

Break down $\csc x \tan x$ into SOHCAHTOA to solve:

$\csc=\frac{\text{hypotenuse}$}{$\text{opposite}$}$ and $\tan=\frac{\text{opposite}$}{$\text{adjacent}$}$ .

Therefore, $\csc x \tan x=\frac{\text{hypotenuse}$}{$\text{opposite}$}*$\frac{\text{opposite}$}{$\text{adjacent}$}$ . Note that the opposite's cancel out, leaving $$\frac{\text{hypotenuse}$}{$\text{adjacent}$}$ , which is the same as $\sec x$ .

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Knew It

Using Sum and Product Identities - Math

Simplify .

To simplify , break them into their SOHCAHTOA parts: . Notice that the opposite's cancel out, leaving .

Simplify .

Break down into SOHCAHTOA to solve: and . Therefore, . Note that the opposite's cancel out, leaving , which is the same as .

Simplify .

To simplify , break them into their SOHCAHTOA parts: . Notice that the opposite's cancel out, leaving .

Simplify .

Break down into SOHCAHTOA to solve: and . Therefore, . Note that the opposite's cancel out, leaving , which is the same as .

Simplify .

To simplify , break them into their SOHCAHTOA parts: . Notice that the opposite's cancel out, leaving .

Simplify .

Break down into SOHCAHTOA to solve: and . Therefore, . Note that the opposite's cancel out, leaving , which is the same as .

Simplify .

To simplify , break them into their SOHCAHTOA parts: . Notice that the opposite's cancel out, leaving .

Simplify .

Break down into SOHCAHTOA to solve: and . Therefore, . Note that the opposite's cancel out, leaving , which is the same as .

Simplify $\sin x\cot x$ .

To simplify $\sin x\cot x$ , break them into their SOHCAHTOA parts:

$$\frac{\text{opposite}$}{$\text{hypotenuse}$}*$\frac{\text{adjacent}$}{$\text{opposite}$}$ .

Notice that the opposite's cancel out, leaving $$\frac{\text{adjacent}$}{$\text{hypotenuse}$}=\cos x$ .

Simplify $\csc x \tan x$ .

Simplify $\sin x\cot x$ .

To simplify $\sin x\cot x$ , break them into their SOHCAHTOA parts:

$$\frac{\text{opposite}$}{$\text{hypotenuse}$}*$\frac{\text{adjacent}$}{$\text{opposite}$}$ .

Notice that the opposite's cancel out, leaving $$\frac{\text{adjacent}$}{$\text{hypotenuse}$}=\cos x$ .

Simplify $\csc x \tan x$ .

Simplify $\sin x\cot x$ .

To simplify $\sin x\cot x$ , break them into their SOHCAHTOA parts:

$$\frac{\text{opposite}$}{$\text{hypotenuse}$}*$\frac{\text{adjacent}$}{$\text{opposite}$}$ .

Notice that the opposite's cancel out, leaving $$\frac{\text{adjacent}$}{$\text{hypotenuse}$}=\cos x$ .

Simplify $\csc x \tan x$ .

Simplify $\sin x\cot x$ .

To simplify $\sin x\cot x$ , break them into their SOHCAHTOA parts:

$$\frac{\text{opposite}$}{$\text{hypotenuse}$}*$\frac{\text{adjacent}$}{$\text{opposite}$}$ .

Notice that the opposite's cancel out, leaving $$\frac{\text{adjacent}$}{$\text{hypotenuse}$}=\cos x$ .

Simplify $\csc x \tan x$ .