Evaluate (sin(theta))/(cos(theta))+(cos(theta))/(sin(theta))=csc(theta)sec(theta)

$\frac{\sin (θ)}{\cos (θ)} + \frac{\cos (θ)}{\sin (θ)} = \csc (θ) \sec (θ)$

Step 1

Simplify the right side.

Simplify $\csc (θ) \sec (θ)$ .

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

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

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

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

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

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

Step 2

Multiply both sides of the equation by $\cos (θ)$ .

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

Step 3

Apply the distributive property.

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

Step 4

Cancel the common factor of $\cos (θ)$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 5

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

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Step 6

Cancel the common factor of $\cos (θ)$ .

Factor $\cos (θ)$ out of $\sin (θ) \cos (θ)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 7

Subtract $\frac{1}{\sin (θ)}$ from both sides of the equation.

$\sin (θ) + \frac{\cos^{2} (θ)}{\sin (θ)} - \frac{1}{\sin (θ)} = 0$

Step 8

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

Combine the numerators over the common denominator.

$\sin (θ) + \frac{\cos^{2} (θ) - 1}{\sin (θ)} = 0$

Reorder $\cos^{2} (θ)$ and $- 1$ .

$\sin (θ) + \frac{- 1 + \cos^{2} (θ)}{\sin (θ)} = 0$

Rewrite $- 1$ as $- 1 (1)$ .

$\sin (θ) + \frac{- 1 (1) + \cos^{2} (θ)}{\sin (θ)} = 0$

Factor $- 1$ out of $\cos^{2} (θ)$ .

$\sin (θ) + \frac{- 1 (1) - 1 (- \cos^{2} (θ))}{\sin (θ)} = 0$

Factor $- 1$ out of $- 1 (1) - 1 (- \cos^{2} (θ))$ .

$\sin (θ) + \frac{- 1 (1 - \cos^{2} (θ))}{\sin (θ)} = 0$

Rewrite $- 1 (1 - \cos^{2} (θ))$ as $- (1 - \cos^{2} (θ))$ .

$\sin (θ) + \frac{- (1 - \cos^{2} (θ))}{\sin (θ)} = 0$

Apply pythagorean identity.

$\sin (θ) + \frac{- \sin^{2} (θ)}{\sin (θ)} = 0$

Cancel the common factor of $\sin^{2} (θ)$ and $\sin (θ)$ .

Factor $\sin (θ)$ out of $- \sin^{2} (θ)$ .

$\sin (θ) + \frac{\sin (θ) (- \sin (θ))}{\sin (θ)} = 0$

Cancel the common factors.

Multiply by $1$ .

$\sin (θ) + \frac{\sin (θ) (- \sin (θ))}{\sin (θ) \cdot 1} = 0$

Cancel the common factor.

$\sin (θ) + \frac{\sin (θ) (- \sin (θ))}{\sin (θ) \cdot 1} = 0$

Rewrite the expression.

$\sin (θ) + \frac{- \sin (θ)}{1} = 0$

Divide $- \sin (θ)$ by $1$ .

$\sin (θ) - \sin (θ) = 0$

Subtract $\sin (θ)$ from $\sin (θ)$ .

$0 = 0$

Step 9

Since $0 = 0$ , the equation will always be true.

Always true

Step 10

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$

Evaluate (sin(theta))/(cos(theta))+(cos(theta))/(sin(theta))=csc(theta)sec(theta)

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