What is the general solution for (cot(θ)+csc(θ))/(sec(θ)+1)=sin(θ) ?

The general solution for (cot(θ)+csc(θ))/(sec(θ)+1)=sin(θ) is θ=0.90455…+2pin,θ=2pi-0.90455…+2pin

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(cot(θ)+csc(θ))/(sec(θ)+1)=sin(θ)

Solution

\frac{cot ( θ ) + csc ( θ )}{sec ( θ ) + 1} = sin (θ)

Solution

θ = 0.90455 \dots + 2 πn, θ = 2 π - 0.90455 \dots + 2 πn

Degrees

θ = 51.82729 \dots^{\circ} + 36 0^{\circ} n, θ = 308.17270 \dots^{\circ} + 36 0^{\circ} n

Solution steps

\frac{cot ( θ ) + csc ( θ )}{sec ( θ ) + 1} = sin (θ)

Subtract

sin (θ)

from both sides

\frac{cot ( θ ) + csc ( θ )}{sec ( θ ) + 1} - sin (θ) = 0

Simplify

\frac{cot ( θ ) + csc ( θ )}{sec ( θ ) + 1} - sin (θ) : \frac{cot ( θ ) + csc ( θ ) - sin ( θ ) ( sec ( θ ) + 1 )}{sec ( θ ) + 1}

\frac{cot ( θ ) + csc ( θ )}{sec ( θ ) + 1} - sin (θ)

Convert element to fraction:

sin (θ) = \frac{sin ( θ ) ( sec ( θ ) + 1 )}{sec ( θ ) + 1}

= \frac{cot ( θ ) + csc ( θ )}{sec ( θ ) + 1} - \frac{sin ( θ ) ( sec ( θ ) + 1 )}{sec ( θ ) + 1}

Since the denominators are equal, combine the fractions:

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

= \frac{cot ( θ ) + csc ( θ ) - sin ( θ ) ( sec ( θ ) + 1 )}{sec ( θ ) + 1}

\frac{cot ( θ ) + csc ( θ ) - sin ( θ ) ( sec ( θ ) + 1 )}{sec ( θ ) + 1} = 0

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

cot (θ) + csc (θ) - sin (θ) (sec (θ) + 1) = 0

Express with sin, cos

cot (θ) + csc (θ) - (1 + sec (θ)) sin (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

\frac{cos ( θ )}{sin ( θ )} + \frac{1}{sin ( θ )} - (1 + \frac{1}{cos ( θ )}) sin (θ) : \frac{cos ( θ ) ( cos ( θ ) + 1 ) - sin ^{2} ( θ ) ( cos ( θ ) + 1 )}{sin ( θ ) cos ( θ )}

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

Combine the fractions

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

Apply rule

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

= \frac{cos ( θ ) + 1}{sin ( θ )}

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

Join

1 + \frac{1}{cos ( θ )} : \frac{cos ( θ ) + 1}{cos ( θ )}

1 + \frac{1}{cos ( θ )}

Convert element to fraction:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) + 1}{cos ( θ )}

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

Multiply

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

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

Multiply fractions:

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

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

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

Least Common Multiplier of

sin (θ), cos (θ) : sin (θ) cos (θ)

sin (θ), cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (θ)

cos (θ)

= sin (θ) cos (θ)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

sin (θ) cos (θ)

For

\frac{cos ( θ ) + 1}{sin ( θ )} :

multiply the denominator and numerator by

cos (θ)

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

For

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

multiply the denominator and numerator by

sin (θ)

\frac{( cos ( θ ) + 1 ) sin ( θ )}{cos ( θ )} = \frac{( cos ( θ ) + 1 ) sin ( θ ) sin ( θ )}{cos ( θ ) sin ( θ )} = \frac{sin ^{2} ( θ ) ( cos ( θ ) + 1 )}{sin ( θ ) cos ( θ )}

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

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

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

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

(1 + cos (θ)) cos (θ) - (1 + cos (θ)) sin^{2} (θ) = 0

Factor

(1 + cos (θ)) cos (θ) - (1 + cos (θ)) sin^{2} (θ) : (1 + cos (θ)) (cos (θ) - sin^{2} (θ))

(1 + cos (θ)) cos (θ) - (1 + cos (θ)) sin^{2} (θ)

Factor out common term

(1 + cos (θ))

= (1 + cos (θ)) (cos (θ) - sin^{2} (θ))

(1 + cos (θ)) (cos (θ) - sin^{2} (θ)) = 0

Solving each part separately

1 + cos (θ) = 0 or cos (θ) - sin^{2} (θ) = 0

1 + cos (θ) = 0 : θ = π + 2 πn

1 + cos (θ) = 0

Move

1

to the right side

1 + cos (θ) = 0

Subtract

1

from both sides

1 + cos (θ) - 1 = 0 - 1

Simplify

cos (θ) = - 1

cos (θ) = - 1

General solutions for

cos (θ) = - 1

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = π + 2 πn

θ = π + 2 πn

cos (θ) - sin^{2} (θ) = 0 : θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (θ) - sin^{2} (θ) = 0

Rewrite using trig identities

cos (θ) - sin^{2} (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= cos (θ) - (1 - cos^{2} (θ))

- (1 - cos^{2} (θ)) : - 1 + cos^{2} (θ)

- (1 - cos^{2} (θ))

Distribute parentheses

= - (1) - (- cos^{2} (θ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (θ)

= cos (θ) - 1 + cos^{2} (θ)

- 1 + cos (θ) + cos^{2} (θ) = 0

Solve by substitution

- 1 + cos (θ) + cos^{2} (θ) = 0

Let:

cos (θ) = u

- 1 + u + u^{2} = 0

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

- 1 + u + u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

u^{2} + u - 1 = 0

Solve with the quadratic formula

u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

1^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

The solutions to the quadratic equation are:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{- 1 + 5}{2}, cos (θ) = \frac{- 1 - 5}{2}

cos (θ) = \frac{- 1 + 5}{2}, cos (θ) = \frac{- 1 - 5}{2}

cos (θ) = \frac{- 1 + 5}{2} : θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (θ) = \frac{- 1 + 5}{2}

Apply trig inverse properties

cos (θ) = \frac{- 1 + 5}{2}

General solutions for

cos (θ) = \frac{- 1 + 5}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (θ) = \frac{- 1 - 5}{2} :

No Solution

cos (θ) = \frac{- 1 - 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

Combine all the solutions

θ = π + 2 πn, θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

Since the equation is undefined for:

π + 2 πn

θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

θ = 0.90455 \dots + 2 πn, θ = 2 π - 0.90455 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (cot(θ)+csc(θ))/(sec(θ)+1)=sin(θ) ?
The general solution for (cot(θ)+csc(θ))/(sec(θ)+1)=sin(θ) is θ=0.90455…+2pin,θ=2pi-0.90455…+2pin