Trigonometry Examples

$\sec (- u)$

Step 1

Set the argument in $\sec (- u)$ equal to $\frac{π}{2} + π n$ to find where the expression is undefined.

$- u = \frac{π}{2} + π n$ , for any integer $n$

Step 2

Divide each term in $- u = \frac{π}{2} + π n$ by $- 1$ and simplify.

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Step 2.1

Divide each term in $- u = \frac{π}{2} + π n$ by $- 1$ .

$\frac{- u}{- 1} = \frac{\frac{π}{2}}{- 1} + \frac{π n}{- 1}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Dividing two negative values results in a positive value.

$\frac{u}{1} = \frac{\frac{π}{2}}{- 1} + \frac{π n}{- 1}$

Step 2.2.2

Divide $u$ by $1$ .

$u = \frac{\frac{π}{2}}{- 1} + \frac{π n}{- 1}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Move the negative one from the denominator of $\frac{\frac{π}{2}}{- 1}$ .

$u = - 1 \cdot \frac{π}{2} + \frac{π n}{- 1}$

Step 2.3.1.2

Rewrite $- 1 \cdot \frac{π}{2}$ as $- \frac{π}{2}$ .

$u = - \frac{π}{2} + \frac{π n}{- 1}$

Step 2.3.1.3

Move the negative one from the denominator of $\frac{π n}{- 1}$ .

$u = - \frac{π}{2} - 1 \cdot (π n)$

Step 2.3.1.4

Rewrite $- 1 \cdot (π n)$ as $- (π n)$ .

$u = - \frac{π}{2} - π n$

Step 3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

${u | u = - \frac{π}{2} - π n}$ $n$ , for any integer $n$

Step 4