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Popular Trigonometry >

tan(90-θ)>1

Solution

tan (9 0^{\circ} - θ) > 1

Solution

πn < θ < \frac{- π + 4 \cdot 9 0 ^{\circ}}{4} + πn

Interval Notation

(πn, \frac{- π + 4 \cdot 9 0 ^{\circ}}{4} + πn)

Decimal

πn < θ < 0.78539 \dots + πn

Solution steps

tan (9 0^{\circ} - θ) > 1

Factor out

- 1

from

9 0^{\circ} - θ : - (- 9 0^{\circ} + θ)

tan (- (- 9 0^{\circ} + θ)) > 1

Use the following identity:

tan (- x) = - tan (x)

- tan (- 9 0^{\circ} + θ) > 1

Multiply both sides by

- 1

- tan (- 9 0^{\circ} + θ) > 1

Multiply both sides by -1 (reverse the inequality)

(- tan (- 9 0^{\circ} + θ)) (- 1) < 1 \cdot (- 1)

Simplify

tan (- 9 0^{\circ} + θ) < - 1

tan (- 9 0^{\circ} + θ) < - 1

tan (x) < a

then

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

- \frac{π}{2} + πn < (- 9 0^{\circ} + θ) < arctan (- 1) + πn

a < u < b

then

a < u and u < b

- \frac{π}{2} + πn < - 9 0^{\circ} + θ and - 9 0^{\circ} + θ < arctan (- 1) + πn

- \frac{π}{2} + πn < - 9 0^{\circ} + θ : θ > \frac{- π + 2 \cdot 9 0 ^{\circ}}{2} + πn

- \frac{π}{2} + πn < - 9 0^{\circ} + θ

Switch sides

- 9 0^{\circ} + θ > - \frac{π}{2} + πn

Move

9 0^{\circ}

to the right side

- 9 0^{\circ} + θ > - \frac{π}{2} + πn

Add

9 0^{\circ}

to both sides

- 9 0^{\circ} + θ + 9 0^{\circ} > - \frac{π}{2} + πn + 9 0^{\circ}

Simplify

θ > - \frac{π}{2} + πn + 9 0^{\circ}

θ > - \frac{π}{2} + πn + 9 0^{\circ}

Simplify

- \frac{π}{2} + 9 0^{\circ} : \frac{- π + 2 \cdot 9 0 ^{\circ}}{2}

- \frac{π}{2} + 9 0^{\circ}

Convert element to fraction:

9 0^{\circ} = \frac{9 0 ^{\circ} \cdot 2}{2}

= - \frac{π}{2} + \frac{9 0 ^{\circ} \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- π + 9 0 ^{\circ} \cdot 2}{2}

θ > \frac{- π + 2 \cdot 9 0 ^{\circ}}{2} + πn

- 9 0^{\circ} + θ < arctan (- 1) + πn : θ < \frac{- π + 4 \cdot 9 0 ^{\circ}}{4} + πn

- 9 0^{\circ} + θ < arctan (- 1) + πn

Simplify

arctan (- 1) + πn : - \frac{π}{4} + πn

arctan (- 1) + πn

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

- 9 0^{\circ} + θ < - \frac{π}{4} + πn

Move

9 0^{\circ}

to the right side

- 9 0^{\circ} + θ < - \frac{π}{4} + πn

Add

9 0^{\circ}

to both sides

- 9 0^{\circ} + θ + 9 0^{\circ} < - \frac{π}{4} + πn + 9 0^{\circ}

Simplify

θ < - \frac{π}{4} + πn + 9 0^{\circ}

θ < - \frac{π}{4} + πn + 9 0^{\circ}

Simplify

- \frac{π}{4} + 9 0^{\circ} : \frac{- π + 4 \cdot 9 0 ^{\circ}}{4}

- \frac{π}{4} + 9 0^{\circ}

Convert element to fraction:

9 0^{\circ} = \frac{9 0 ^{\circ} \cdot 4}{4}

= - \frac{π}{4} + \frac{9 0 ^{\circ} \cdot 4}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{- π + 9 0 ^{\circ} \cdot 4}{4}

θ < \frac{- π + 4 \cdot 9 0 ^{\circ}}{4} + πn

Combine the intervals

θ > \frac{- π + 2 \cdot 9 0 ^{\circ}}{2} + πn and θ < \frac{- π + 4 \cdot 9 0 ^{\circ}}{4} + πn

Merge Overlapping Intervals

πn < θ < \frac{- π + 4 \cdot 9 0 ^{\circ}}{4} + πn

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