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Popular >

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

Solution

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

Solution

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

Interval Notation

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

Decimal Notation

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

Solution steps

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

Let:

u = tan (x)

\frac{1 + u}{1 - u} > 0

\frac{1 + u}{1 - u} > 0 : - 1 < u < 1

\frac{1 + u}{1 - u} > 0

Identify the intervals

Find the signs of the factors of

\frac{1 + u}{1 - u}

Find the signs of

1 + u

1 + u = 0 : u = - 1

1 + u = 0

Move

1

to the right side

1 + u = 0

Subtract

1

from both sides

1 + u - 1 = 0 - 1

Simplify

u = - 1

u = - 1

1 + u < 0 : u < - 1

1 + u < 0

Move

1

to the right side

1 + u < 0

Subtract

1

from both sides

1 + u - 1 < 0 - 1

Simplify

u < - 1

u < - 1

1 + u > 0 : u > - 1

1 + u > 0

Move

1

to the right side

1 + u > 0

Subtract

1

from both sides

1 + u - 1 > 0 - 1

Simplify

u > - 1

u > - 1

Find the signs of

1 - u

1 - u = 0 : u = 1

1 - u = 0

Move

1

to the right side

1 - u = 0

Subtract

1

from both sides

1 - u - 1 = 0 - 1

Simplify

- u = - 1

- u = - 1

Divide both sides by

- 1

- u = - 1

Divide both sides by

- 1

\frac{- u}{- 1} = \frac{- 1}{- 1}

Simplify

u = 1

u = 1

1 - u < 0 : u > 1

1 - u < 0

Move

1

to the right side

1 - u < 0

Subtract

1

from both sides

1 - u - 1 < 0 - 1

Simplify

- u < - 1

- u < - 1

Multiply both sides by

- 1

- u < - 1

Multiply both sides by -1 (reverse the inequality)

(- u) (- 1) > (- 1) (- 1)

Simplify

u > 1

u > 1

1 - u > 0 : u < 1

1 - u > 0

Move

1

to the right side

1 - u > 0

Subtract

1

from both sides

1 - u - 1 > 0 - 1

Simplify

- u > - 1

- u > - 1

Multiply both sides by

- 1

- u > - 1

Multiply both sides by -1 (reverse the inequality)

(- u) (- 1) < (- 1) (- 1)

Simplify

u < 1

u < 1

Find singularity points

Find the zeros of the denominator

1 - u : u = 1

1 - u = 0

Move

1

to the right side

1 - u = 0

Subtract

1

from both sides

1 - u - 1 = 0 - 1

Simplify

- u = - 1

- u = - 1

Divide both sides by

- 1

- u = - 1

Divide both sides by

- 1

\frac{- u}{- 1} = \frac{- 1}{- 1}

Simplify

u = 1

u = 1

Summarize in a table:

1 + u 1 - u \frac{1 + u}{1 - u} u < - 1 - + - u = - 1 0 + 0 - 1 < u < 1 + + + u = 1 + 0 Undefined u > 1 + - -

Identify the intervals that satisfy the required condition:

> 0

- 1 < u < 1

- 1 < u < 1

- 1 < u < 1

Substitute back

u = tan (x)

- 1 < tan (x) < 1

a < u < b

then

a < u and u < b

- 1 < tan (x) and tan (x) < 1

- 1 < tan (x) : - \frac{π}{4} + πn < x < \frac{π}{2} + πn

- 1 < tan (x)

Switch sides

tan (x) > - 1

tan (x) > a

then

arctan (a) + πn < x < \frac{π}{2} + πn

arctan (- 1) + πn < x < \frac{π}{2} + πn

Simplify

arctan (- 1) : - \frac{π}{4}

arctan (- 1)

Use the following property:

arctan (- x) = - arctan (x)

arctan (- 1) = - arctan (1)

= - arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

arctan (1)

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

= - \frac{π}{4}

- \frac{π}{4} + πn < x < \frac{π}{2} + πn

tan (x) < 1 : - \frac{π}{2} + πn < x < \frac{π}{4} + πn

tan (x) < 1

tan (x) < a

then

- \frac{π}{2} + πn < x < arctan (a) + πn

- \frac{π}{2} + πn < x < arctan (1) + πn

Simplify

arctan (1) : \frac{π}{4}

arctan (1)

Use the following trivial identity:

arctan (1) = \frac{π}{4}

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= \frac{π}{4}

- \frac{π}{2} + πn < x < \frac{π}{4} + πn

Combine the intervals

- \frac{π}{4} + πn < x < \frac{π}{2} + πn and - \frac{π}{2} + πn < x < \frac{π}{4} + πn

Merge Overlapping Intervals

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

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