Using Sum and Product Identities - Math

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Question

Simplify $\sin x\cot x$ .

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 $\csc x \tan x$ .

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 $\sec x$ .

Compare your answer with the correct one above

Question

Simplify $\sin x\cot x$ .

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 $\frac{\text{adjacent}}{\text{hypotenuse}}=\cos x$ .

Compare your answer with the correct one above

Question

Simplify $\csc x \tan x$ .

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 $\sec x$ .

Compare your answer with the correct one above

Question

Simplify $\sin x\cot x$ .

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 $\frac{\text{adjacent}}{\text{hypotenuse}}=\cos x$ .

Compare your answer with the correct one above

Question

Simplify $\csc x \tan x$ .

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 $\sec x$ .

Compare your answer with the correct one above

Question

Simplify $\sin x\cot x$ .

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 $\frac{\text{adjacent}}{\text{hypotenuse}}=\cos x$ .

Compare your answer with the correct one above

Question

Simplify $\csc x \tan x$ .

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 $\sec x$ .

Compare your answer with the correct one above

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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 $\csc x \tan x$ .