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Popular Trigonometry >

(cos(x)(1+tan(x)))/(cos(x)(1-tan(x)))>0

Solution

\frac{cos ( x ) ( 1 + tan ( x ) )}{cos ( x ) ( 1 - tan ( x ) )} > 0

Solution

πn \leq x < \frac{π}{4} + πn or \frac{3 π}{4} + πn < x \leq π + πn

Interval Notation

[πn, \frac{π}{4} + πn) \cup (\frac{3 π}{4} + πn, π + πn]

Decimal

πn \leq x < 0.78539 \dots + πn or 2.35619 \dots + πn < x \leq 3.14159 \dots + πn

Solution steps

\frac{cos ( x ) ( 1 + tan ( x ) )}{cos ( x ) ( 1 - tan ( x ) )} > 0

Periodicity of

\frac{cos ( x ) ( 1 + tan ( x ) )}{cos ( x ) ( 1 - tan ( x ) )} : π

\frac{cos ( x ) ( 1 + tan ( x ) )}{cos ( x ) ( 1 - tan ( x ) )}

is composed of the following functions and periods:

cos (x)

with periodicity of

2 π

The compound periodicity is:

= π

Express with sin, cos

\frac{cos ( x ) ( 1 + tan ( x ) )}{cos ( x ) ( 1 - tan ( x ) )} > 0

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

\frac{cos ( x ) ( 1 + \frac{s i n ( x )}{c o s ( x )} )}{cos ( x ) ( 1 - \frac{s i n ( x )}{c o s ( x )} )} > 0

\frac{cos ( x ) ( 1 + \frac{s i n ( x )}{c o s ( x )} )}{cos ( x ) ( 1 - \frac{s i n ( x )}{c o s ( x )} )} > 0

Simplify

\frac{cos ( x ) ( 1 + \frac{s i n ( x )}{c o s ( x )} )}{cos ( x ) ( 1 - \frac{s i n ( x )}{c o s ( x )} )} : \frac{cos ( x ) + sin ( x )}{cos ( x ) - sin ( x )}

\frac{cos ( x ) ( 1 + \frac{s i n ( x )}{c o s ( x )} )}{cos ( x ) ( 1 - \frac{s i n ( x )}{c o s ( x )} )}

Cancel the common factor:

cos (x)

= \frac{1 + \frac{s i n ( x )}{c o s ( x )}}{1 - \frac{s i n ( x )}{c o s ( x )}}

Join

1 - \frac{sin ( x )}{cos ( x )} : \frac{cos ( x ) - sin ( x )}{cos ( x )}

1 - \frac{sin ( x )}{cos ( x )}

Convert element to fraction:

1 = \frac{1 cos ( x )}{cos ( x )}

= \frac{1 \cdot cos ( x )}{cos ( x )} - \frac{sin ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot cos ( x ) - sin ( x )}{cos ( x )}

Multiply:

1 \cdot cos (x) = cos (x)

= \frac{cos ( x ) - sin ( x )}{cos ( x )}

= \frac{1 + \frac{s i n ( x )}{c o s ( x )}}{\frac{c o s ( x ) - s i n ( x )}{c o s ( x )}}

Join

1 + \frac{sin ( x )}{cos ( x )} : \frac{cos ( x ) + sin ( x )}{cos ( x )}

1 + \frac{sin ( x )}{cos ( x )}

Convert element to fraction:

1 = \frac{1 cos ( x )}{cos ( x )}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Multiply:

1 \cdot cos (x) = cos (x)

= \frac{cos ( x ) + sin ( x )}{cos ( x )}

= \frac{\frac{c o s ( x ) + s i n ( x )}{c o s ( x )}}{\frac{c o s ( x ) - s i n ( x )}{c o s ( x )}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{( cos ( x ) + sin ( x ) ) cos ( x )}{cos ( x ) ( cos ( x ) - sin ( x ) )}

Cancel the common factor:

cos (x)

= \frac{cos ( x ) + sin ( x )}{cos ( x ) - sin ( x )}

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

Find the zeroes and undifined points of

\frac{cos ( x ) + sin ( x )}{cos ( x ) - sin ( x )}

for

0 \leq x < π

To find the zeroes, set the inequality to zero

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

\frac{cos ( x ) + sin ( x )}{cos ( x ) - sin ( x )} = 0, 0 \leq x < π : x = \frac{3 π}{4}

\frac{cos ( x ) + sin ( x )}{cos ( x ) - sin ( x )} = 0, 0 \leq x < π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos (x) + sin (x) = 0

Rewrite using trig identities

cos (x) + sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{cos ( x ) + sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 + tan (x) = 0

1 + tan (x) = 0

Move

1

to the right side

1 + tan (x) = 0

Subtract

1

from both sides

1 + tan (x) - 1 = 0 - 1

Simplify

tan (x) = - 1

tan (x) = - 1

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Solutions for the range

0 \leq x < π

x = \frac{3 π}{4}

Find the undefined points:

x = \frac{π}{4}

Find the zeros of the denominator

cos (x) - sin (x) = 0

Rewrite using trig identities

cos (x) - sin (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{cos ( x ) - sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

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

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 - tan (x) = 0

1 - tan (x) = 0

Move

1

to the right side

1 - tan (x) = 0

Subtract

1

from both sides

1 - tan (x) - 1 = 0 - 1

Simplify

- tan (x) = - 1

- tan (x) = - 1

Divide both sides by

- 1

- tan (x) = - 1

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solutions for the range

0 \leq x < π

x = \frac{π}{4}

\frac{π}{4}, \frac{3 π}{4}

Identify the intervals

0 < x < \frac{π}{4}, \frac{π}{4} < x < \frac{3 π}{4}, \frac{3 π}{4} < x < π

Summarize in a table:

cos (x) + sin (x) cos (x) - sin (x) \frac{c o s ( x ) + s i n ( x )}{c o s ( x ) - s i n ( x )} x = 0 + + + 0 < x < \frac{π}{4} + + + x = \frac{π}{4} + 0 Undefined \frac{π}{4} < x < \frac{3 π}{4} + - - x = \frac{3 π}{4} 0 - 0 \frac{3 π}{4} < x < π - - + x = π - - +

Identify the intervals that satisfy the required condition:

> 0

x = 0 or 0 < x < \frac{π}{4} or \frac{3 π}{4} < x < π or x = π

Merge Overlapping Intervals

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < π or x = π

The union of two intervals is the set of numbers which are in either interval

x = 0

0 < x < \frac{π}{4}

0 \leq x < \frac{π}{4}

The union of two intervals is the set of numbers which are in either interval

0 \leq x < \frac{π}{4}

\frac{3 π}{4} < x < π

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < π

The union of two intervals is the set of numbers which are in either interval

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x < π

x = π

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x \leq π

0 \leq x < \frac{π}{4} or \frac{3 π}{4} < x \leq π

Apply the periodicity of

\frac{cos ( x ) ( 1 + tan ( x ) )}{cos ( x ) ( 1 - tan ( x ) )}

πn \leq x < \frac{π}{4} + πn or \frac{3 π}{4} + πn < x \leq π + πn

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