What is the general solution for sec(θ)csc(θ)=1 ?

The general solution for sec(θ)csc(θ)=1 is No Solution for θ\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sec(θ)csc(θ)=1

Solution

sec (θ) csc (θ) = 1

Solution

No Solution for θ \in R

Solution steps

sec (θ) csc (θ) = 1

Subtract

1

from both sides

sec (θ) csc (θ) - 1 = 0

Express with sin, cos

- 1 + csc (θ) sec (θ)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= - 1 + \frac{1}{sin ( θ )} sec (θ)

Use the basic trigonometric identity:

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

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

Simplify

- 1 + \frac{1}{sin ( θ )} \cdot \frac{1}{cos ( θ )} : \frac{- sin ( θ ) cos ( θ ) + 1}{sin ( θ ) cos ( θ )}

- 1 + \frac{1}{sin ( θ )} \cdot \frac{1}{cos ( θ )}

\frac{1}{sin ( θ )} \cdot \frac{1}{cos ( θ )} = \frac{1}{sin ( θ ) cos ( θ )}

\frac{1}{sin ( θ )} \cdot \frac{1}{cos ( θ )}

Multiply fractions:

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

= \frac{1 \cdot 1}{sin ( θ ) cos ( θ )}

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{sin ( θ ) cos ( θ )}

= - 1 + \frac{1}{sin ( θ ) cos ( θ )}

Convert element to fraction:

1 = \frac{1 sin ( θ ) cos ( θ )}{sin ( θ ) cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{- sin ( θ ) cos ( θ ) + 1}{sin ( θ ) cos ( θ )}

= \frac{- sin ( θ ) cos ( θ ) + 1}{sin ( θ ) cos ( θ )}

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

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

1 - cos (θ) sin (θ) = 0

Rewrite using trig identities

1 - cos (θ) sin (θ)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= 1 - \frac{sin ( 2 θ )}{2}

1 - \frac{sin ( 2 θ )}{2} = 0

Move

1

to the right side

1 - \frac{sin ( 2 θ )}{2} = 0

Subtract

1

from both sides

1 - \frac{sin ( 2 θ )}{2} - 1 = 0 - 1

Simplify

- \frac{sin ( 2 θ )}{2} = - 1

- \frac{sin ( 2 θ )}{2} = - 1

Multiply both sides by

2

- \frac{sin ( 2 θ )}{2} = - 1

Multiply both sides by

2

2 (- \frac{sin ( 2 θ )}{2}) = 2 (- 1)

Simplify

- sin (2 θ) = - 2

- sin (2 θ) = - 2

Divide both sides by

- 1

- sin (2 θ) = - 2

Divide both sides by

- 1

\frac{- sin ( 2 θ )}{- 1} = \frac{- 2}{- 1}

Simplify

sin (2 θ) = 2

sin (2 θ) = 2

- 1 \leq sin (x) \leq 1

No Solution for θ \in R

Popular Examples

prove cos(a+270)=sin(a)prove

cos (a + 27 0^{\circ}) = sin (a)

1+sin(θ)=2-sin(θ)

1 + sin (θ) = 2 - sin (θ)

-4cos^2(θ)+cos(θ)=9cos(θ)+3

- 4 cos^{2} (θ) + cos (θ) = 9 cos (θ) + 3

3+4cot(x)=5csc(x)

3 + 4 cot (x) = 5 csc (x)

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

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

Frequently Asked Questions (FAQ)

What is the general solution for sec(θ)csc(θ)=1 ?
The general solution for sec(θ)csc(θ)=1 is No Solution for θ\in\mathbb{R}