Precalculus Examples

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

Step 1

Start on the left side.

$\sin (θ) \csc (θ) - \cot^{2} (θ) \sin^{2} (θ)$

Step 2

Apply Pythagorean identity in reverse.

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

Step 3

Convert to sines and cosines.

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Step 3.1

Apply the reciprocal identity to $\csc (θ)$ .

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

Step 3.2

Apply the reciprocal identity to $\csc (θ)$ .

$\sin (θ) \frac{1}{\sin (θ)} - ({(\frac{1}{\sin (θ)})}^{2} - 1) \sin^{2} (θ)$

Step 3.3

Apply the product rule to $\frac{1}{\sin (θ)}$ .

$\sin (θ) \frac{1}{\sin (θ)} - (\frac{1^{2}}{\sin^{2} (θ)} - 1) \sin^{2} (θ)$

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

Cancel the common factor of $\sin (θ)$ .

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Step 4.1.1.1

Cancel the common factor.

$\sin (θ) \frac{1}{\sin (θ)} - (\frac{1^{2}}{{\sin (θ)}^{2}} - 1) {\sin (θ)}^{2}$

Step 4.1.1.2

Rewrite the expression.

$1 - (\frac{1^{2}}{{\sin (θ)}^{2}} - 1) {\sin (θ)}^{2}$

Step 4.1.2

One to any power is one.

$1 - (\frac{1}{{\sin (θ)}^{2}} - 1) {\sin (θ)}^{2}$

Step 4.1.3

Apply the distributive property.

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

Step 4.1.4

Multiply $- 1$ by $- 1$ .

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

Step 4.1.5

Apply the distributive property.

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

Step 4.1.6

Cancel the common factor of ${\sin (θ)}^{2}$ .

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Step 4.1.6.1

Move the leading negative in $- \frac{1}{{\sin (θ)}^{2}}$ into the numerator.

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

Step 4.1.6.2

Cancel the common factor.

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

Step 4.1.6.3

Rewrite the expression.

$1 - 1 + 1 {\sin (θ)}^{2}$

Step 4.1.7

Multiply ${\sin (θ)}^{2}$ by $1$ .

$1 - 1 + {\sin (θ)}^{2}$

Step 4.2

Subtract $1$ from $1$ .

$0 + {\sin (θ)}^{2}$

Step 4.3

Add $0$ and ${\sin (θ)}^{2}$ .

$\sin^{2} (θ)$

Step 5

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

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