Precalculus Examples

$\sin (θ) + \cos (θ) \cot (θ) = \csc (θ)$

Step 1

Start on the left side.

$\sin (θ) + \cos (θ) \cot (θ)$

Step 2

Simplify each term.

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Step 2.1

Rewrite $\cot (θ)$ in terms of sines and cosines.

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Raise $\cos (θ)$ to the power of $1$ .

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

Step 2.2.3

Raise $\cos (θ)$ to the power of $1$ .

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

Step 2.2.4

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

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

Step 2.2.5

Add $1$ and $1$ .

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

Step 3

Apply Pythagorean identity in reverse.

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

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

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

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 $b = \sin (θ)$ .

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

Step 4.2

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

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

Step 4.3

Combine the numerators over the common denominator.

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

Step 4.4

Simplify the numerator.

$\frac{1}{\sin (θ)}$

Step 5

Rewrite $\frac{1}{\sin (θ)}$ as $\csc (θ)$ .

$\csc (θ)$

Step 6

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

$\sin (θ) + \cos (θ) \cot (θ) = \csc (θ)$ is an identity