Trigonometry Examples

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

Step 1

Start on the left side.

$\csc^{2} (θ) \cdot \tan^{2} (θ) - 1$

Step 2

Apply Pythagorean identity in reverse.

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

Step 3

Convert to sines and cosines.

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Step 3.1

Write $\cot (θ)$ in sines and cosines using the quotient identity.

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

Step 3.2

Write $\tan (θ)$ in sines and cosines using the quotient identity.

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

Step 3.3

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

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

Step 3.4

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

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

Step 4

Simplify.

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Step 4.1

Simplify each term.

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Step 4.1.1

Apply the distributive property.

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

Step 4.1.2

Multiply $\frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}}$ by $1$ .

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

Step 4.1.3

Combine.

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

Step 4.1.4

Simplify each term.

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Step 4.1.4.1

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

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Step 4.1.4.1.1

Cancel the common factor.

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

Step 4.1.4.1.2

Rewrite the expression.

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

Step 4.1.4.2

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

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Step 4.1.4.2.1

Cancel the common factor.

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

Step 4.1.4.2.2

Rewrite the expression.

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

Step 4.2

Subtract $1$ from $1$ .

$\frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}} + 0$

Step 4.3

Add $\frac{{\sin (θ)}^{2}}{{\cos (θ)}^{2}}$ and $0$ .

$\frac{\sin^{2} (θ)}{\cos^{2} (θ)}$

Step 5

Rewrite $\frac{\sin^{2} (θ)}{\cos^{2} (θ)}$ as $\tan^{2} (θ)$ .

$\tan^{2} (θ)$

Step 6

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

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