Precalculus Examples

$1 + \tan^{2} (- θ) = \sec^{2} (θ)$

Step 1

Start on the left side.

$1 + \tan^{2} (- θ)$

Step 2

Since $\tan (- θ)$ is an odd function, rewrite $\tan (- θ)$ as $- \tan (θ)$ .

$1 + {(- \tan (θ))}^{2}$

Step 3

Convert to sines and cosines.

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Step 3.1

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

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

Step 3.2

Simplify.

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Step 3.2.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.2.1.1

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

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

Step 3.2.1.2

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

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

Step 3.2.2

Raise $- 1$ to the power of $2$ .

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

Step 3.2.3

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

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

Step 4

Apply Pythagorean identity in reverse.

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

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Rewrite $1$ as $1^{2}$ .

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

Step 5.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = \cos (θ)$ .

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

Step 5.2

Write $1$ as a fraction with a common denominator.

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

Step 5.3

Combine the numerators over the common denominator.

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

Step 5.4

Simplify the numerator.

$\frac{1}{\cos^{2} (θ)}$

Step 6

Rewrite $\frac{1}{\cos^{2} (θ)}$ as $\sec^{2} (θ)$ .

$\sec^{2} (θ)$

Step 7

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

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