Precalculus Examples

$(\csc (θ) + 1) (\csc (θ) - 1) = \cot^{2} (θ)$

Step 1

Start on the left side.

$(\csc (θ) + 1) (\csc (θ) - 1)$

Step 2

Apply the distributive property.

$(\csc (θ) + 1) \csc (θ) + (\csc (θ) + 1) \cdot - 1$

Step 3

Simplify.

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Step 3.1

Simplify each term.

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Step 3.1.1

Apply the distributive property.

$\csc (θ) \csc (θ) + 1 \csc (θ) + (\csc (θ) + 1) \cdot - 1$

Step 3.1.2

Multiply $\csc (θ) \csc (θ)$ .

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Step 3.1.2.1

Raise $\csc (θ)$ to the power of $1$ .

${\csc (θ)}^{1} \csc (θ) + 1 \csc (θ) + (\csc (θ) + 1) \cdot - 1$

Step 3.1.2.2

Raise $\csc (θ)$ to the power of $1$ .

${\csc (θ)}^{1} {\csc (θ)}^{1} + 1 \csc (θ) + (\csc (θ) + 1) \cdot - 1$

Step 3.1.2.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${\csc (θ)}^{1 + 1} + 1 \csc (θ) + (\csc (θ) + 1) \cdot - 1$

Step 3.1.2.4

Add $1$ and $1$ .

${\csc (θ)}^{2} + 1 \csc (θ) + (\csc (θ) + 1) \cdot - 1$

Step 3.1.3

Multiply $\csc (θ)$ by $1$ .

${\csc (θ)}^{2} + \csc (θ) + (\csc (θ) + 1) \cdot - 1$

Step 3.1.4

Apply the distributive property.

${\csc (θ)}^{2} + \csc (θ) + \csc (θ) \cdot - 1 + 1 \cdot - 1$

Step 3.1.5

Move $- 1$ to the left of $\csc (θ)$ .

${\csc (θ)}^{2} + \csc (θ) - 1 \cdot \csc (θ) + 1 \cdot - 1$

Step 3.1.6

Multiply $- 1$ by $1$ .

${\csc (θ)}^{2} + \csc (θ) - 1 \cdot \csc (θ) - 1$

Step 3.1.7

Rewrite $- 1 \csc (θ)$ as $- \csc (θ)$ .

${\csc (θ)}^{2} + \csc (θ) - \csc (θ) - 1$

Step 3.2

Subtract $\csc (θ)$ from $\csc (θ)$ .

${\csc (θ)}^{2} + 0 - 1$

Step 3.3

Add ${\csc (θ)}^{2}$ and $0$ .

$\csc^{2} (θ) - 1$

Step 4

Apply pythagorean identity.

$\cot^{2} (θ)$

Step 5

Because the two sides have been shown to be equivalent, the equation is an identity.

$(\csc (θ) + 1) (\csc (θ) - 1) = \cot^{2} (θ)$ is an identity