simplifying trig identities worksheet
1. cos x tan x
2. cos x csc x
3. sin x sec x
4. tan x csc x
5. tan²x - sec²x
6. $\frac{sec x}{csc x}$
7. sin x + cot x cos x
8. cos²x(1 + tan²x)
9. $\frac{sec x - cos x}{sin x}$
10. $\frac{cot x}{csc x - sin x}$
11. $\frac{sin x sec x}{tan x}$
12. cos³x + sin²x cos x
13. $\frac{1 + cos x}{1 + sec x}$
14. $\frac{tan x}{sec(x)}$
15. $\frac{sec²x - 1}{sec²x}$
16. $\frac{sin x}{csc x} + \frac{cos x}{sec x}$

Question

Accepted Answer

Let's take problem 1: $\boldsymbol{\cos x 	imes 	an x}$ to simplify.

# Explanation:
## Step1: Recall the definition of $	an x$
We know that $	an x = \frac{\sin x}{\cos x}$. Substitute this into the expression:
$\cos x 	imes 	an x = \cos x 	imes \frac{\sin x}{\cos x}$

## Step2: Simplify the expression
The $\cos x$ in the numerator and denominator cancels out:
$\cos x 	imes \frac{\sin x}{\cos x} = \sin x$

# Answer:
$\sin x$

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Let's take problem 2: $\boldsymbol{\cos x 	imes \csc x}$ to simplify.

# Explanation:
## Step1: Recall the definition of $\csc x$
We know that $\csc x = \frac{1}{\sin x}$. Substitute this into the expression:
$\cos x 	imes \csc x = \cos x 	imes \frac{1}{\sin x}$

## Step2: Simplify the expression
We know that $\frac{\cos x}{\sin x} = \cot x$, so:
$\cos x 	imes \frac{1}{\sin x} = \frac{\cos x}{\sin x} = \cot x$

# Answer:
$\cot x$

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Let's take problem 3: $\boldsymbol{\sin x 	imes \sec x}$ to simplify.

# Explanation:
## Step1: Recall the definition of $\sec x$
We know that $\sec x = \frac{1}{\cos x}$. Substitute this into the expression:
$\sin x 	imes \sec x = \sin x 	imes \frac{1}{\cos x}$

## Step2: Simplify the expression
We know that $\frac{\sin x}{\cos x} = 	an x$, so:
$\sin x 	imes \frac{1}{\cos x} = \frac{\sin x}{\cos x} = 	an x$

# Answer:
$	an x$

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Let's take problem 4: $\boldsymbol{	an x 	imes \csc x}$ to simplify.

# Explanation:
## Step1: Recall the definitions of $	an x$ and $\csc x$
We know that $	an x = \frac{\sin x}{\cos x}$ and $\csc x = \frac{1}{\sin x}$. Substitute these into the expression:
$	an x 	imes \csc x = \frac{\sin x}{\cos x} 	imes \frac{1}{\sin x}$

## Step2: Simplify the expression
The $\sin x$ in the numerator and denominator cancels out:
$\frac{\sin x}{\cos x} 	imes \frac{1}{\sin x} = \frac{1}{\cos x} = \sec x$

# Answer:
$\sec x$

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Let's take problem 5: $\boldsymbol{	an^2 x - \sec^2 x}$ to simplify.

# Explanation:
## Step1: Recall the Pythagorean identity
We know the Pythagorean identity: $	an^2 x + 1 = \sec^2 x$, which can be rearranged to $	an^2 x - \sec^2 x = -1$.

## Step2: Verify the identity
Starting from $	an^2 x + 1 = \sec^2 x$, subtract $\sec^2 x$ and 1 from both sides:
$	an^2 x - \sec^2 x = -1$

# Answer:
$-1$

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Let's take problem 6: $\boldsymbol{\frac{\sec x}{\csc x}}$ to simplify.

# Explanation:
## Step1: Recall the definitions of $\sec x$ and $\csc x$
We know that $\sec x = \frac{1}{\cos x}$ and $\csc x = \frac{1}{\sin x}$. Substitute these into the expression:
$\frac{\sec x}{\csc x} = \frac{\frac{1}{\cos x}}{\frac{1}{\sin x}}$

## Step2: Simplify the complex fraction
Dividing by a fraction is the same as multiplying by its reciprocal:
$\frac{\frac{1}{\cos x}}{\frac{1}{\sin x}} = \frac{1}{\cos x} 	imes \frac{\sin x}{1} = \frac{\sin x}{\cos x} = 	an x$

# Answer:
$	an x$

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Let's take problem 7: $\boldsymbol{\sin x + \cot x 	imes \cos x}$ to simplify.

# Explanation:
## Step1: Recall the definition of $\cot x$
We know that $\cot x = \frac{\cos x}{\sin x}$. Substitute this into the expression:
$\sin x + \cot x 	imes \cos x = \sin x + \frac{\cos x}{\sin x} 	imes \cos x$

## Step2: Simplify the second term
$\frac{\cos x}{\sin x} 	imes \cos x = \frac{\cos^2 x}{\sin x}$

## Step3: Combine the terms over a common denominator
$\sin x + \frac{\cos^2 x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x}$

## Step4: Use the Pythagorean identity
We know that $\sin^2 x + \cos^2 x = 1$, so:
$\frac{\sin^2 x + \cos^2 x}{\sin x} = \frac{1}{\sin x} = \csc x$

# Answer:
$\csc x$

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Let's take problem 8: $\boldsymbol{\cos^2 x (1 + 	an^2 x)}$ to simplify.

# Explanation:
## Step1: Recall the Pythagorean identity
We know the Pythagorean identity: $1 + 	an^2 x = \sec^2 x$. Substitute this into the expression:
$\cos^2 x (1 + 	an^2 x) = \cos^2 x 	imes \sec^2 x$

## Step2: Recall the definition of $\sec x$
We know that $\sec x = \frac{1}{\cos x}$, so $\sec^2 x = \frac{1}{\cos^2 x}$. Substitute this into the expression:
$\cos^2 x 	imes \frac{1}{\cos^2 x} = 1$

# Answer:
$1$

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Let's take problem 9: $\boldsymbol{\frac{\sec x - \cos x}{\sin x}}$ to simplify.

# Explanation:
## Step1: Recall the definition of $\sec x$
We know that $\sec x = \frac{1}{\cos x}$. Substitute this into the numerator:
$\sec x - \cos x = \frac{1}{\cos x} - \cos x$

## Step2: Combine the terms in the numerator over a common denominator
$\frac{1}{\cos x} - \cos x = \frac{1 - \cos^2 x}{\cos x}$

## Step3: Use the Pythagorean identity
We know that $1 - \cos^2 x = \sin^2 x$, so:
$\frac{1 - \cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x}$

## Step4: Divide by $\sin x$
$\frac{\frac{\sin^2 x}{\cos x}}{\sin x} = \frac{\sin^2 x}{\cos x} 	imes \frac{1}{\sin x} = \frac{\sin x}{\cos x} = 	an x$

# Answer:
$	an x$

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Let's take problem 10: $\boldsymbol{\frac{\cot x}{\csc x - \sin x}}$ to simplify.

# Explanation:
## Step1: Recall the definitions of $\cot x$ and $\csc x$
We know that $\cot x = \frac{\cos x}{\sin x}$ and $\csc x = \frac{1}{\sin x}$. Substitute these into the expression:
$\frac{\cot x}{\csc x - \sin x} = \frac{\frac{\cos x}{\sin x}}{\frac{1}{\sin x} - \sin x}$

## Step2: Combine the terms in the denominator over a common denominator
$\frac{1}{\sin x} - \sin x = \frac{1 - \sin^2 x}{\sin x}$

## Step3: Use the Pythagorean identity
We know that $1 - \sin^2 x = \cos^2 x$, so:
$\frac{1 - \sin^2 x}{\sin x} = \frac{\cos^2 x}{\sin x}$

## Step4: Simplify the complex fraction
$\frac{\frac{\cos x}{\sin x}}{\frac{\cos^2 x}{\sin x}} = \frac{\cos x}{\sin x} 	imes \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} = \sec x$

# Answer:
$\sec x$

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Let's take problem 11: $\boldsymbol{\frac{\sin x 	imes \sec x}{	an x}}$ to simplify.

# Explanation:
## Step1: Recall the definitions of $\sec x$ and $	an x$
We know that $\sec x = \frac{1}{\cos x}$ and $	an x = \frac{\sin x}{\cos x}$. Substitute these into the expression:
$\frac{\sin x 	imes \sec x}{	an x} = \frac{\sin x 	imes \frac{1}{\cos x}}{\frac{\sin x}{\cos x}}$

## Step2: Simplify the complex fraction
Multiply the numerator and denominator by $\cos x$ to eliminate the fractions:
$\frac{\sin x 	imes \frac{1}{\cos x} 	imes \cos x}{\frac{\sin x}{\cos x} 	imes \cos x} = \frac{\sin x}{\sin x} = 1$

# Answer:
$1$

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Let's take problem 12: $\boldsymbol{\cos^3 x + \sin^2 x 	imes \cos x}$ to simplify.

# Explanation:
## Step1: Factor out $\cos x$ from both terms
$\cos^3 x + \sin^2 x 	imes \cos x = \cos x (\cos^2 x + \sin^2 x)$

## Step2: Use the Pythagorean identity
We know that $\cos^2 x + \sin^2 x = 1$, so:
$\cos x (\cos^2 x + \sin^2 x) = \cos x 	imes 1 = \cos x$

# Answer:
$\cos x$

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Let's take problem 13: $\boldsymbol{\frac{1 + \cos x}{1 + \sec x}}$ to simplify.

# Explanation:
## Step1: Recall the definition of $\sec x$
We know that $\sec x = \frac{1}{\cos x}$. Substitute this into the denominator:
$\frac{1 + \cos x}{1 + \sec x} = \frac{1 + \cos x}{1 + \frac{1}{\cos x}}$

## Step2: Combine the terms in the denominator over a common denominator
$1 + \frac{1}{\cos x} = \frac{\cos x + 1}{\cos x}$

## Step3: Simplify the complex fraction
$\frac{1 + \cos x}{\frac{\cos x + 1}{\cos x}} = (1 + \cos x) 	imes \frac{\cos x}{\cos x + 1} = \cos x$

# Answer:
$\cos x$

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Let's take problem 14: $\boldsymbol{\frac{	an x}{\sec x}}$ to simplify.

# Explanation:
## Step1: Recall the definitions of $	an x$ and $\sec x$
We know that $	an x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. Substitute these into the expression:
$\frac{	an x}{\sec x} = \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}$

## Step2: Simplify the complex fraction
Multiply the numerator and denominator by $\cos x$ to eliminate the fractions:
$\frac{\frac{\sin x}{\cos x} 	imes \cos x}{\frac{1}{\cos x} 	imes \cos x} = \frac{\sin x}{1} = \sin x$

# Answer:
$\sin x$

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Let's take problem 15: $\boldsymbol{\frac{\sec^2 x - 1}{\sec^2 x}}$ to simplify.

# Explanation:
## Step1: Recall the Pythagorean identity
We know that $	an^2 x + 1 = \sec^2 x$, so $\sec^2 x - 1 = 	an^2 x$. Substitute this into the numerator:
$\frac{\sec^2 x - 1}{\sec^2 x} = \frac{	an^2 x}{\sec^2 x}$

## Step2: Recall the definitions of $	an x$ and $\sec x$
We know that $	an x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$, so $	an^2 x = \frac{\sin^2 x}{\cos^2 x}$ and $\sec^2 x = \frac{1}{\cos^2 x}$. Substitute these into the expression:
$\frac{\frac{\sin^2 x}{\cos^2 x}}{\frac{1}{\cos^2 x}} = \frac{\sin^2 x}{\cos^2 x} 	imes \frac{\cos^2 x}{1} = \sin^2 x$

# Answer:
$\sin^2 x$

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Let's take problem 16: $\boldsymbol{\frac{\sin x}{\csc x} + \frac{\cos x}{\sec x}}$ to simplify.

# Explanation:
## Step1: Recall the definitions of $\csc x$ and $\sec x$
We know that $\csc x = \frac{1}{\sin x}$ and $\sec x = \frac{1}{\cos x}$. Substitute these into the expression:
$\frac{\sin x}{\csc x} + \frac{\cos x}{\sec x} = \frac{\sin x}{\frac{1}{\sin x}} + \frac{\cos x}{\frac{1}{\cos x}}$

## Step2: Simplify each term
$\frac{\sin x}{\frac{1}{\sin x}} = \sin^2 x$ and $\frac{\cos x}{\frac{1}{\cos x}} = \cos^2 x$

## Step3: Combine the terms
$\sin^2 x + \cos^2 x = 1$ (by the Pythagorean identity)

# Answer:
$1$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the definition of $\tan x$

Step2: Simplify the expression

Step1: Recall the definition of $\csc x$

Step2: Simplify the expression

Step1: Recall the definition of $\sec x$

Step2: Simplify the expression

Answer: