Determine if True tan(theta)+cot(theta)=csc(theta)sec(theta)

$\tan (θ) + \cot (θ) = \csc (θ) \sec (θ)$

Step 1

Start on the left side.

$\tan (θ) + \cot (θ)$

Step 2

Convert to sines and cosines.

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

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

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

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

Step 3

Add fractions.

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

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

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

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

Write each expression with a common denominator of $\cos (θ) \sin (θ)$ , by multiplying each by an appropriate factor of $1$ .

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

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

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

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

Reorder the factors of $\sin (θ) \cos (θ)$ .

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

Combine the numerators over the common denominator.

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

Step 4

Simplify each term.

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

Step 5

Apply pythagorean identity.

$\frac{1}{\cos (θ) \sin (θ)}$

Step 6

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

$\csc (θ) \sec (θ)$

Step 7

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

$\tan (θ) + \cot (θ) = \csc (θ) \sec (θ)$ is an identity