Question

name:____ date:____ m2 precalculus: section 8.4 \verifying trigonometric identities\ leveled practice 1. level i a. csc a tan a = sec a c. cos θ sec θ - cos³θ = sin²θ b. $\frac{cot\beta}{csc\beta}=cos\beta$ d. cot y sec y sin y = 1 2. level ii: a. $\frac{\tan^{2}\theta}{1 + \tan^{2}\theta}=sin^{2}\theta$ c. cos θ sin²θ + cos³θ b. sin θ+cos θ cot θ = csc θ d. (sec θ - tan θ)(csc θ + 1)=cot θ

Explanation:

Step1: Rewrite in terms of sine and cosine

Recall that $\csc A=\frac{1}{\sin A}$, $\tan A = \frac{\sin A}{\cos A}$, and $\sec A=\frac{1}{\cos A}$. Then $\csc A\tan A=\frac{1}{\sin A}\cdot\frac{\sin A}{\cos A}$.
$\frac{1}{\sin A}\cdot\frac{\sin A}{\cos A}=\frac{1}{\cos A}$
## Step2: Identify the result
Since $\frac{1}{\cos A}=\sec A$, the identity is verified.
- # Answer:
The identity $\csc A\tan A = \sec A$ is true.
2. **b. Verify $\frac{\cot\beta}{\csc\beta}=\cos\beta$**:
- # Explanation:
## Step1: Rewrite cotangent and cosecant
Recall that $\cot\beta=\frac{\cos\beta}{\sin\beta}$ and $\csc\beta=\frac{1}{\sin\beta}$. Then $\frac{\cot\beta}{\csc\beta}=\frac{\frac{\cos\beta}{\sin\beta}}{\frac{1}{\sin\beta}}$.
$\frac{\frac{\cos\beta}{\sin\beta}}{\frac{1}{\sin\beta}}=\frac{\cos\beta}{\sin\beta}\cdot\sin\beta$
## Step2: Simplify the expression
$\frac{\cos\beta}{\sin\beta}\cdot\sin\beta=\cos\beta$
- # Answer:
The identity $\frac{\cot\beta}{\csc\beta}=\cos\beta$ is true.
3. **c. Verify $\cos\theta\sec\theta-\cos^{2}\theta=\sin^{2}\theta$**:
- # Explanation:
## Step1: Simplify $\cos\theta\sec\theta$
Recall that $\sec\theta=\frac{1}{\cos\theta}$, so $\cos\theta\sec\theta=\cos\theta\cdot\frac{1}{\cos\theta}=1$.
The left - hand side becomes $1 - \cos^{2}\theta$.
## Step2: Use the Pythagorean identity
By the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$, we can rewrite $1-\cos^{2}\theta$ as $\sin^{2}\theta$.
- # Answer:
The identity $\cos\theta\sec\theta-\cos^{2}\theta=\sin^{2}\theta$ is true.
4. **d. Verify $\cot y\sec y\sin y = 1$**:
- # Explanation:
## Step1: Rewrite cotangent and secant
Recall that $\cot y=\frac{\cos y}{\sin y}$ and $\sec y=\frac{1}{\cos y}$. Then $\cot y\sec y\sin y=\frac{\cos y}{\sin y}\cdot\frac{1}{\cos y}\cdot\sin y$.
$\frac{\cos y}{\sin y}\cdot\frac{1}{\cos y}\cdot\sin y = 1$
- # Answer:
The identity $\cot y\sec y\sin y=1$ is true.
5. **2. a. Verify $\frac{\tan^{2}\theta}{1 + \tan^{2}\theta}=\sin^{2}\theta$**:
- # Explanation:
## Step1: Recall the Pythagorean identity $1+\tan^{2}\theta=\sec^{2}\theta$
So the left - hand side becomes $\frac{\tan^{2}\theta}{\sec^{2}\theta}$.
Since $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\sec\theta=\frac{1}{\cos\theta}$, we have $\frac{\tan^{2}\theta}{\sec^{2}\theta}=\frac{\frac{\sin^{2}\theta}{\cos^{2}\theta}}{\frac{1}{\cos^{2}\theta}}$.
$\frac{\frac{\sin^{2}\theta}{\cos^{2}\theta}}{\frac{1}{\cos^{2}\theta}}=\frac{\sin^{2}\theta}{\cos^{2}\theta}\cdot\cos^{2}\theta$
## Step2: Simplify the expression
$\frac{\sin^{2}\theta}{\cos^{2}\theta}\cdot\cos^{2}\theta=\sin^{2}\theta$
- # Answer:
The identity $\frac{\tan^{2}\theta}{1+\tan^{2}\theta}=\sin^{2}\theta$ is true.
6. **b. Verify $\sin\theta+\cos\theta\cot\theta=\csc\theta$**:
- # Explanation:
## Step1: Rewrite cotangent
Recall that $\cot\theta=\frac{\cos\theta}{\sin\theta}$. Then $\cos\theta\cot\theta=\cos\theta\cdot\frac{\cos\theta}{\sin\theta}=\frac{\cos^{2}\theta}{\sin\theta}$.
The left - hand side is $\sin\theta+\frac{\cos^{2}\theta}{\sin\theta}=\frac{\sin^{2}\theta+\cos^{2}\theta}{\sin\theta}$.
## Step2: Use the Pythagorean identity
Since $\sin^{2}\theta+\cos^{2}\theta = 1$, $\frac{\sin^{2}\theta+\cos^{2}\theta}{\sin\theta}=\frac{1}{\sin\theta}=\csc\theta$.
- # Answer:
The identity $\sin\theta+\cos\theta\cot\theta=\csc\theta$ is true.
7. **c. Simplify $\cos\theta\sin^{2}\theta+\cos^{3}\theta$**:
- # Explanation:
## Step1: Factor out $\cos\theta$
$\cos\theta\sin^{2}\theta+\cos^{3}\theta=\cos\theta(\sin^{2}\theta+\cos^{2}\theta)$
## Step2: Use the Pythagorean identity
Since $\sin^{2}\theta+\cos^{2}\theta = 1$, $\cos\theta(\sin^{2}\theta+\cos^{2}\theta)=\cos\theta$.
- # Answer:
$\cos\theta\sin^{2}\theta+\cos^{3}\theta=\cos\theta$
8. **d. Verify $(\sec\theta-\tan\theta)(\csc\theta + 1)=\cot\theta$**:
- # Explanation:
## Step1: Expand the left - hand side
$(\sec\theta-\tan\theta)(\csc\theta + 1)=\sec\theta\csc\theta+\sec\theta-\tan\theta\csc\theta-\tan\theta$.
Recall that $\sec\theta=\frac{1}{\cos\theta}$, $\csc\theta=\frac{1}{\sin\theta}$, and $\tan\theta=\frac{\sin\theta}{\cos\theta}$.
$\sec\theta\csc\theta=\frac{1}{\cos\theta\sin\theta}$, $\tan\theta\csc\theta=\frac{\sin\theta}{\cos\theta}\cdot\frac{1}{\sin\theta}=\frac{1}{\cos\theta}$.
The left - hand side becomes $\frac{1}{\cos\theta\sin\theta}+\frac{1}{\cos\theta}-\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}=\frac{1 - \sin^{2}\theta}{\cos\theta\sin\theta}$.
## Step2: Use the Pythagorean identity
Since $1-\sin^{2}\theta=\cos^{2}\theta$, $\frac{1 - \sin^{2}\theta}{\cos\theta\sin\theta}=\frac{\cos^{2}\theta}{\cos\theta\sin\theta}=\frac{\cos\theta}{\sin\theta}=\cot\theta$.
- # Answer:
The identity $(\sec\theta-\tan\theta)(\csc\theta + 1)=\cot\theta$ is true.

Answer:

1. a. Verify $\csc A\tan A=\sec A$:

• # Explanation:

Step1: Rewrite in terms of sine and cosine

Recall that $\csc A=\frac{1}{\sin A}$, $\tan A = \frac{\sin A}{\cos A}$, and $\sec A=\frac{1}{\cos A}$. Then $\csc A\tan A=\frac{1}{\sin A}\cdot\frac{\sin A}{\cos A}$.
$\frac{1}{\sin A}\cdot\frac{\sin A}{\cos A}=\frac{1}{\cos A}$

Step2: Identify the result

Since $\frac{1}{\cos A}=\sec A$, the identity is verified.

• # Answer:

The identity $\csc A\tan A = \sec A$ is true.

1. b. Verify $\frac{\cot\beta}{\csc\beta}=\cos\beta$:

• # Explanation:

Step1: Rewrite cotangent and cosecant

Recall that $\cot\beta=\frac{\cos\beta}{\sin\beta}$ and $\csc\beta=\frac{1}{\sin\beta}$. Then $\frac{\cot\beta}{\csc\beta}=\frac{\frac{\cos\beta}{\sin\beta}}{\frac{1}{\sin\beta}}$.
$\frac{\frac{\cos\beta}{\sin\beta}}{\frac{1}{\sin\beta}}=\frac{\cos\beta}{\sin\beta}\cdot\sin\beta$

Step2: Simplify the expression

$\frac{\cos\beta}{\sin\beta}\cdot\sin\beta=\cos\beta$

• # Answer:

The identity $\frac{\cot\beta}{\csc\beta}=\cos\beta$ is true.

1. c. Verify $\cos\theta\sec\theta-\cos^{2}\theta=\sin^{2}\theta$:

• # Explanation:

Step1: Simplify $\cos\theta\sec\theta$

Recall that $\sec\theta=\frac{1}{\cos\theta}$, so $\cos\theta\sec\theta=\cos\theta\cdot\frac{1}{\cos\theta}=1$.
The left - hand side becomes $1 - \cos^{2}\theta$.

Step2: Use the Pythagorean identity

By the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$, we can rewrite $1-\cos^{2}\theta$ as $\sin^{2}\theta$.

• # Answer:

The identity $\cos\theta\sec\theta-\cos^{2}\theta=\sin^{2}\theta$ is true.

1. d. Verify $\cot y\sec y\sin y = 1$:

• # Explanation:

Step1: Rewrite cotangent and secant

Recall that $\cot y=\frac{\cos y}{\sin y}$ and $\sec y=\frac{1}{\cos y}$. Then $\cot y\sec y\sin y=\frac{\cos y}{\sin y}\cdot\frac{1}{\cos y}\cdot\sin y$.
$\frac{\cos y}{\sin y}\cdot\frac{1}{\cos y}\cdot\sin y = 1$

• # Answer:

The identity $\cot y\sec y\sin y=1$ is true.

1. 2. a. Verify $\frac{\tan^{2}\theta}{1 + \tan^{2}\theta}=\sin^{2}\theta$:

• # Explanation:

Step1: Recall the Pythagorean identity $1+\tan^{2}\theta=\sec^{2}\theta$

So the left - hand side becomes $\frac{\tan^{2}\theta}{\sec^{2}\theta}$.
Since $\tan\theta=\frac{\sin\theta}{\cos\theta}$ and $\sec\theta=\frac{1}{\cos\theta}$, we have $\frac{\tan^{2}\theta}{\sec^{2}\theta}=\frac{\frac{\sin^{2}\theta}{\cos^{2}\theta}}{\frac{1}{\cos^{2}\theta}}$.
$\frac{\frac{\sin^{2}\theta}{\cos^{2}\theta}}{\frac{1}{\cos^{2}\theta}}=\frac{\sin^{2}\theta}{\cos^{2}\theta}\cdot\cos^{2}\theta$

Step2: Simplify the expression

$\frac{\sin^{2}\theta}{\cos^{2}\theta}\cdot\cos^{2}\theta=\sin^{2}\theta$

• # Answer:

The identity $\frac{\tan^{2}\theta}{1+\tan^{2}\theta}=\sin^{2}\theta$ is true.

1. b. Verify $\sin\theta+\cos\theta\cot\theta=\csc\theta$:

• # Explanation:

Step1: Rewrite cotangent

Recall that $\cot\theta=\frac{\cos\theta}{\sin\theta}$. Then $\cos\theta\cot\theta=\cos\theta\cdot\frac{\cos\theta}{\sin\theta}=\frac{\cos^{2}\theta}{\sin\theta}$.
The left - hand side is $\sin\theta+\frac{\cos^{2}\theta}{\sin\theta}=\frac{\sin^{2}\theta+\cos^{2}\theta}{\sin\theta}$.

Step2: Use the Pythagorean identity

Since $\sin^{2}\theta+\cos^{2}\theta = 1$, $\frac{\sin^{2}\theta+\cos^{2}\theta}{\sin\theta}=\frac{1}{\sin\theta}=\csc\theta$.

• # Answer:

The identity $\sin\theta+\cos\theta\cot\theta=\csc\theta$ is true.

1. c. Simplify $\cos\theta\sin^{2}\theta+\cos^{3}\theta$:

• # Explanation:

Step1: Factor out $\cos\theta$

$\cos\theta\sin^{2}\theta+\cos^{3}\theta=\cos\theta(\sin^{2}\theta+\cos^{2}\theta)$

Step2: Use the Pythagorean identity

Since $\sin^{2}\theta+\cos^{2}\theta = 1$, $\cos\theta(\sin^{2}\theta+\cos^{2}\theta)=\cos\theta$.

• # Answer:

$\cos\theta\sin^{2}\theta+\cos^{3}\theta=\cos\theta$

1. d. Verify $(\sec\theta-\tan\theta)(\csc\theta + 1)=\cot\theta$:

• # Explanation:

Step1: Expand the left - hand side

$(\sec\theta-\tan\theta)(\csc\theta + 1)=\sec\theta\csc\theta+\sec\theta-\tan\theta\csc\theta-\tan\theta$.
Recall that $\sec\theta=\frac{1}{\cos\theta}$, $\csc\theta=\frac{1}{\sin\theta}$, and $\tan\theta=\frac{\sin\theta}{\cos\theta}$.
$\sec\theta\csc\theta=\frac{1}{\cos\theta\sin\theta}$, $\tan\theta\csc\theta=\frac{\sin\theta}{\cos\theta}\cdot\frac{1}{\sin\theta}=\frac{1}{\cos\theta}$.
The left - hand side becomes $\frac{1}{\cos\theta\sin\theta}+\frac{1}{\cos\theta}-\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}=\frac{1 - \sin^{2}\theta}{\cos\theta\sin\theta}$.

Step2: Use the Pythagorean identity

Since $1-\sin^{2}\theta=\cos^{2}\theta$, $\frac{1 - \sin^{2}\theta}{\cos\theta\sin\theta}=\frac{\cos^{2}\theta}{\cos\theta\sin\theta}=\frac{\cos\theta}{\sin\theta}=\cot\theta$.

• # Answer:

The identity $(\sec\theta-\tan\theta)(\csc\theta + 1)=\cot\theta$ is true.