Find the Determinant [[1,0,sin(x)],[1,0,cos(x)],[1,sec(x),-sin(x)]]

$[\begin{array}{c} 1 & 0 & \sin (x) \\ 1 & 0 & \cos (x) \\ 1 & \sec (x) & - \sin (x) \end{array}]$

Set up the determinant by breaking it into smaller components.

$0 | \begin{array}{c} 1 & \cos (x) \\ 1 & - \sin (x) \end{array} | + 0 | \begin{array}{c} 1 & \sin (x) \\ 1 & - \sin (x) \end{array} | - \sec (x) | \begin{array}{c} 1 & \sin (x) \\ 1 & \cos (x) \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$0 + 0 | \begin{array}{c} 1 & \sin (x) \\ 1 & - \sin (x) \end{array} | - \sec (x) | \begin{array}{c} 1 & \sin (x) \\ 1 & \cos (x) \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$0 + 0 - \sec (x) | \begin{array}{c} 1 & \sin (x) \\ 1 & \cos (x) \end{array} |$

The determinant of $- \sec (x) | \begin{array}{c} 1 & \sin (x) \\ 1 & \cos (x) \end{array} |$ is $- 1 + \tan (x)$ .

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The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$0 + 0 - \sec (x) ((1) (\cos (x)) - 1 \sin (x))$

Simplify the determinant.

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

$0 + 0 - \frac{1}{\cos (x)} \cdot ((1) (\cos (x)) - 1 \sin (x))$

Simplify each term.

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Multiply $\cos (x)$ by $1$ .

$0 + 0 - \frac{1}{\cos (x)} \cdot (\cos (x) - 1 \sin (x))$

Rewrite $- 1 \sin (x)$ as $- \sin (x)$ .

$0 + 0 - \frac{1}{\cos (x)} \cdot (\cos (x) - \sin (x))$

Simplify terms.

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Apply the distributive property.

$0 + 0 - \frac{1}{\cos (x)} \cdot \cos (x) - \frac{1}{\cos (x)} \cdot (- \sin (x))$

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

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Move the leading negative in $- \frac{1}{\cos (x)}$ into the numerator.

$0 + 0 + \frac{- 1}{\cos (x)} \cdot \cos (x) - \frac{1}{\cos (x)} \cdot (- \sin (x))$

Cancel the common factor.

$0 + 0 + \frac{- 1}{\cos (x)} \cdot \cos (x) - \frac{1}{\cos (x)} \cdot (- \sin (x))$

Rewrite the expression.

$0 + 0 - 1 - \frac{1}{\cos (x)} \cdot (- \sin (x))$

Multiply $- \frac{1}{\cos (x)} (- \sin (x))$ .

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Multiply $- 1$ by $- 1$ .

$0 + 0 - 1 + 1 (\frac{1}{\cos (x)}) \cdot \sin (x)$

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

$0 + 0 - 1 + \frac{1}{\cos (x)} \cdot \sin (x)$

Combine $\frac{1}{\cos (x)}$ and $\sin (x)$ .

$0 + 0 - 1 + \frac{\sin (x)}{\cos (x)}$

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$0 + 0 - 1 + \tan (x)$

Combine the opposite terms in $0 + 0 - 1 + \tan (x)$ .

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Add $0$ and $0$ .

$0 - 1 + \tan (x)$

Subtract $1$ from $0$ .

$- 1 + \tan (x)$

Find the Determinant [[1,0,sin(x)],[1,0,cos(x)],[1,sec(x),-sin(x)]]