exploration: writing other trigonometric identities
work with a partner. the trigonometric identity you derived in exploration 1 is called
a pythagorean identity. there are two other pythagorean identities. to derive them,
recall the four relationships:
$\tan \theta = \frac{\sin \theta}{\cos \theta}$ $\cot \theta = \frac{\cos \theta}{\sin \theta}$
$\sec \theta = \frac{1}{\cos \theta}$ $\csc \theta = \frac{1}{\sin \theta}$
a. divide each side of the pythagorean identity you derived in exploration 1 by
$\cos^2 \theta$ and simplify. what do you observe?
b. divide each side of the pythagorean identity you derived in exploration 1 by
$\sin^2 \theta$ and simplify. what do you observe?

Question

Accepted Answer

# Explanation:
## Step1: Recall the Pythagorean identity
The Pythagorean identity derived in Exploration 1 is $\sin^{2}\theta+\cos^{2}\theta = 1$.
## Step2: Divide by $\cos^{2}\theta$ (for part a)
Divide each term of $\sin^{2}\theta+\cos^{2}\theta = 1$ by $\cos^{2}\theta$:
$$
\begin{align*}
\frac{\sin^{2}\theta}{\cos^{2}\theta}+\frac{\cos^{2}\theta}{\cos^{2}\theta}&=\frac{1}{\cos^{2}\theta}\
\tan^{2}\theta + 1&=\sec^{2}\theta
\end{align*}
$$
## Step3: Divide by $\sin^{2}\theta$ (for part b)
Divide each term of $\sin^{2}\theta+\cos^{2}\theta = 1$ by $\sin^{2}\theta$:
$$
\begin{align*}
\frac{\sin^{2}\theta}{\sin^{2}\theta}+\frac{\cos^{2}\theta}{\sin^{2}\theta}&=\frac{1}{\sin^{2}\theta}\
1+\cot^{2}\theta&=\csc^{2}\theta
\end{align*}
$$

# Answer:
a. When dividing $\sin^{2}\theta+\cos^{2}\theta = 1$ by $\cos^{2}\theta$, we get $\tan^{2}\theta + 1=\sec^{2}\theta$.
b. When dividing $\sin^{2}\theta+\cos^{2}\theta = 1$ by $\sin^{2}\theta$, we get $1+\cot^{2}\theta=\csc^{2}\theta$.

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the Pythagorean identity

Step2: Divide by \(\cos^{2}\theta\) (for part a)

Step3: Divide by \(\sin^{2}\theta\) (for part b)

Answer: