Trigonometry Examples

$\frac{1}{1 + \sin (x)} + \frac{1}{1 - \sin (x)} = 2 \sec^{2} (x)$

Step 1

Subtract $2 \sec^{2} (x)$ from both sides of the equation.

$\frac{1}{1 + \sin (x)} + \frac{1}{1 - \sin (x)} - 2 \sec^{2} (x) = 0$

Step 2

Simplify $\frac{1}{1 + \sin (x)} + \frac{1}{1 - \sin (x)} - 2 \sec^{2} (x)$ .

Tap for more steps...

Step 2.1

Simplify each term.

Tap for more steps...

Step 2.1.1

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

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

Step 2.1.2

Apply the product rule to $\frac{1}{\cos (x)}$ .

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

Step 2.1.3

One to any power is one.

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

Step 2.1.4

Combine $- 2$ and $\frac{1}{\cos^{2} (x)}$ .

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

Step 2.1.5

Move the negative in front of the fraction.

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

Step 2.2

To write $\frac{1}{1 + \sin (x)}$ as a fraction with a common denominator, multiply by $\frac{1 - \sin (x)}{1 - \sin (x)}$ .

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

Step 2.3

To write $\frac{1}{1 - \sin (x)}$ as a fraction with a common denominator, multiply by $\frac{1 + \sin (x)}{1 + \sin (x)}$ .

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

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Reorder the factors of $(1 - \sin (x)) (1 + \sin (x))$ .

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

Step 2.5

Combine the numerators over the common denominator.

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

Step 2.6

Combine the opposite terms in $\frac{1 - \sin (x) + 1 + \sin (x)}{(1 + \sin (x)) (1 - \sin (x))} - \frac{2}{\cos^{2} (x)}$ .

Tap for more steps...

Step 2.6.1

Add $- \sin (x)$ and $\sin (x)$ .

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

Step 2.6.2

Add $1$ and $0$ .

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

Step 2.7

Add $1$ and $1$ .

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

Step 2.8

Simplify each term.

Tap for more steps...

Step 2.8.1

Multiply by $1$ .

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

Step 2.8.2

Separate fractions.

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

Step 2.8.3

Convert from $\frac{1}{\cos^{2} (x)}$ to $\sec^{2} (x)$ .

$\frac{2}{(1 + \sin (x)) (1 - \sin (x))} - (\frac{2}{1} \sec^{2} (x)) = 0$

Step 2.8.4

Divide $2$ by $1$ .

$\frac{2}{(1 + \sin (x)) (1 - \sin (x))} - (2 \sec^{2} (x)) = 0$

Step 2.8.5

Multiply $2$ by $- 1$ .

$\frac{2}{(1 + \sin (x)) (1 - \sin (x))} - 2 \sec^{2} (x) = 0$

Step 3

Set the denominator in $\frac{2}{(1 + \sin (x)) (1 - \sin (x))}$ equal to $0$ to find where the expression is undefined.

$(1 + \sin (x)) (1 - \sin (x)) = 0$

Step 4

Solve for $x$ .

Tap for more steps...

Step 4.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$1 + \sin (x) = 0$

$1 - \sin (x) = 0$

Step 4.2

Set $1 + \sin (x)$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 4.2.1

Set $1 + \sin (x)$ equal to $0$ .

$1 + \sin (x) = 0$

Step 4.2.2

Solve $1 + \sin (x) = 0$ for $x$ .

Tap for more steps...

Step 4.2.2.1

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 4.2.2.2

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (- 1)$

Step 4.2.2.3

Simplify the right side.

Tap for more steps...

Step 4.2.2.3.1

The exact value of $\arcsin (- 1)$ is $- \frac{π}{2}$ .

$x = - \frac{π}{2}$

Step 4.2.2.4

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 4.2.2.5

Simplify the expression to find the second solution.

Tap for more steps...

Step 4.2.2.5.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 4.2.2.5.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 4.2.2.6

Find the period of $\sin (x)$ .

Tap for more steps...

Step 4.2.2.6.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.2.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.2.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 4.2.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.2.7

Add $2 π$ to every negative angle to get positive angles.

Tap for more steps...

Step 4.2.2.7.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 4.2.2.7.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 4.2.2.7.3

Combine fractions.

Tap for more steps...

Step 4.2.2.7.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 4.2.2.7.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 4.2.2.7.4

Simplify the numerator.

Tap for more steps...

Step 4.2.2.7.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 4.2.2.7.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 4.2.2.7.5

List the new angles.

$x = \frac{3 π}{2}$

Step 4.2.2.8

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{3 π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 4.3

Set $1 - \sin (x)$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 4.3.1

Set $1 - \sin (x)$ equal to $0$ .

$1 - \sin (x) = 0$

Step 4.3.2

Solve $1 - \sin (x) = 0$ for $x$ .

Tap for more steps...

Step 4.3.2.1

Subtract $1$ from both sides of the equation.

$- \sin (x) = - 1$

Step 4.3.2.2

Divide each term in $- \sin (x) = - 1$ by $- 1$ and simplify.

Tap for more steps...

Step 4.3.2.2.1

Divide each term in $- \sin (x) = - 1$ by $- 1$ .

$\frac{- \sin (x)}{- 1} = \frac{- 1}{- 1}$

Step 4.3.2.2.2

Simplify the left side.

Tap for more steps...

Step 4.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{\sin (x)}{1} = \frac{- 1}{- 1}$

Step 4.3.2.2.2.2

Divide $\sin (x)$ by $1$ .

$\sin (x) = \frac{- 1}{- 1}$

Step 4.3.2.2.3

Simplify the right side.

Tap for more steps...

Step 4.3.2.2.3.1

Divide $- 1$ by $- 1$ .

$\sin (x) = 1$

Step 4.3.2.3

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (1)$

Step 4.3.2.4

Simplify the right side.

Tap for more steps...

Step 4.3.2.4.1

The exact value of $\arcsin (1)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 4.3.2.5

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - \frac{π}{2}$

Step 4.3.2.6

Simplify $π - \frac{π}{2}$ .

Tap for more steps...

Step 4.3.2.6.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = π \cdot \frac{2}{2} - \frac{π}{2}$

Step 4.3.2.6.2

Combine fractions.

Tap for more steps...

Step 4.3.2.6.2.1

Combine $π$ and $\frac{2}{2}$ .

$x = \frac{π \cdot 2}{2} - \frac{π}{2}$

Step 4.3.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 4.3.2.6.3

Simplify the numerator.

Tap for more steps...

Step 4.3.2.6.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 4.3.2.6.3.2

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Step 4.3.2.7

Find the period of $\sin (x)$ .

Tap for more steps...

Step 4.3.2.7.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.3.2.7.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.3.2.7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 4.3.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.3.2.8

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

Step 4.4

The final solution is all the values that make $(1 + \sin (x)) (1 - \sin (x)) = 0$ true.

$x = \frac{3 π}{2} + 2 π n, \frac{π}{2} + 2 π n$ , for any integer $n$

Step 4.5

Consolidate the answers.

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

Step 5

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 $0$ .

${x | x = \frac{π}{2} + π n}$ $n$ , for any integer $n$

Step 6