Precalculus Examples

$\sin (θ) \tan (θ) = \sec (θ) - \cos (θ)$

Step 1

Start on the right side.

$\sec (θ) - \cos (θ)$

Step 2

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

$\frac{1}{\cos (θ)} - \cos (θ)$

Step 3

Write $- \cos (θ)$ as a fraction with denominator $1$ .

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

Step 4

Add fractions.

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Step 4.1

To write $\frac{- \cos (θ)}{1}$ as a fraction with a common denominator, multiply by $\frac{\cos (θ)}{\cos (θ)}$ .

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

Step 4.2

Multiply $\frac{- \cos (θ)}{1}$ by $\frac{\cos (θ)}{\cos (θ)}$ .

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

Step 4.3

Combine the numerators over the common denominator.

$\frac{1 - \cos (θ) \cos (θ)}{\cos (θ)}$

Step 5

Multiply $- \cos (θ) \cos (θ)$ .

$\frac{1 - \cos^{2} (θ)}{\cos (θ)}$

Step 6

Apply pythagorean identity.

$\frac{\sin^{2} (θ)}{\cos (θ)}$

Step 7

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

$\sin (θ) \tan (θ)$

Step 8

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

$\sin (θ) \tan (θ) = \sec (θ) - \cos (θ)$ is an identity