Trigonometry Examples

\tan (x) \sin (x) + \cos (x) = \sec (x)

$\tan (x) \sin (x) + \cos (x) = \sec (x)$

Step 1

Start on the left side.

\tan (x) \sin (x) + \cos (x)

$\tan (x) \sin (x) + \cos (x)$

Step 2

Simplify each term.

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Step 2.1

Rewrite

\tan (x)

$\tan (x)$ in terms of sines and cosines.

\frac{\sin (x)}{\cos (x)} \sin (x) + \cos (x)

$\frac{\sin (x)}{\cos (x)} \sin (x) + \cos (x)$

Step 2.2

Multiply

\frac{\sin (x)}{\cos (x)} \sin (x)

$\frac{\sin (x)}{\cos (x)} \sin (x)$ .

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Step 2.2.1

Combine

\frac{\sin (x)}{\cos (x)}

$\frac{\sin (x)}{\cos (x)}$ and

\sin (x)

$\sin (x)$ .

\frac{\sin (x) \sin (x)}{\cos (x)} + \cos (x)

$\frac{\sin (x) \sin (x)}{\cos (x)} + \cos (x)$

Step 2.2.2

Raise

\sin (x)

$\sin (x)$ to the power of

1

$1$ .

\frac{\sin^{1} (x) \sin (x)}{\cos (x)} + \cos (x)

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

Step 2.2.3

Raise

\sin (x)

$\sin (x)$ to the power of

1

$1$ .

\frac{\sin^{1} (x) \sin^{1} (x)}{\cos (x)} + \cos (x)

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

Step 2.2.4

Use the power rule

a^{m} a^{n} = a^{m + n}

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

\frac{{\sin (x)}^{1 + 1}}{\cos (x)} + \cos (x)

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

Step 2.2.5

Add

1

$1$ and

1

$1$ .

\frac{\sin^{2} (x)}{\cos (x)} + \cos (x)

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

\frac{\sin^{2} (x)}{\cos (x)} + \cos (x)

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

\frac{\sin^{2} (x)}{\cos (x)} + \cos (x)

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

Step 3

Apply Pythagorean identity in reverse.

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

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Rewrite

1

$1$ as

1^{2}

$1^{2}$ .

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

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

Step 4.1.2

Since both terms are perfect squares, factor using the difference of squares formula,

a^{2} - b^{2} = (a + b) (a - b)

$a^{2} - b^{2} = (a + b) (a - b)$ where

a = 1

$a = 1$ and

b = \cos (x)

$b = \cos (x)$ .

\frac{(1 + \cos (x)) (1 - \cos (x))}{\cos (x)} + \cos (x)

$\frac{(1 + \cos (x)) (1 - \cos (x))}{\cos (x)} + \cos (x)$

\frac{(1 + \cos (x)) (1 - \cos (x))}{\cos (x)} + \cos (x)

$\frac{(1 + \cos (x)) (1 - \cos (x))}{\cos (x)} + \cos (x)$

Step 4.2

To write

\cos (x)

$\cos (x)$ as a fraction with a common denominator, multiply by

\frac{\cos (x)}{\cos (x)}

$\frac{\cos (x)}{\cos (x)}$ .

\frac{(1 + \cos (x)) (1 - \cos (x))}{\cos (x)} + \frac{\cos (x) \cos (x)}{\cos (x)}

$\frac{(1 + \cos (x)) (1 - \cos (x))}{\cos (x)} + \frac{\cos (x) \cos (x)}{\cos (x)}$

Step 4.3

Combine the numerators over the common denominator.

\frac{(1 + \cos (x)) (1 - \cos (x)) + \cos (x) \cos (x)}{\cos (x)}

$\frac{(1 + \cos (x)) (1 - \cos (x)) + \cos (x) \cos (x)}{\cos (x)}$

Step 4.4

Simplify the numerator.

\frac{1}{\cos (x)}

$\frac{1}{\cos (x)}$

\frac{1}{\cos (x)}

$\frac{1}{\cos (x)}$

Step 5

Rewrite

\frac{1}{\cos (x)}

$\frac{1}{\cos (x)}$ as

\sec (x)

$\sec (x)$ .

\sec (x)

$\sec (x)$

Step 6

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

\tan (x) \sin (x) + \cos (x) = \sec (x)

$\tan (x) \sin (x) + \cos (x) = \sec (x)$ is an identity