Trigonometry Examples

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

Step 1

Start on the left side.

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

Step 2

Simplify each term.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Raise $\sin (x)$ to the power of $1$ .

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

Step 2.2.3

Raise $\sin (x)$ to the power of $1$ .

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

Step 2.2.4

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

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

Step 2.2.5

Add $1$ and $1$ .

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

Step 3

Apply Pythagorean identity in reverse.

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

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

Step 4.1.2

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

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

Step 4.2

To write $\cos (x)$ as a fraction with a common denominator, multiply by $\frac{\cos (x)}{\cos (x)}$ .

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

Step 4.3

Combine the numerators over the common denominator.

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

Step 4.4

Simplify the numerator.

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

Step 5

Rewrite $\frac{1}{\cos (x)}$ as $\sec (x)$ .

$\sec (x)$

Step 6

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

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