What is the general solution for cos(θ)tan(θ)+cot(θ)=csc(θ) ?

The general solution for cos(θ)tan(θ)+cot(θ)=csc(θ) is No Solution for θ\in\mathbb{R}

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cos(θ)tan(θ)+cot(θ)=csc(θ)

Solution

cos (θ) tan (θ) + cot (θ) = csc (θ)

Solution

No Solution for θ \in R

Solution steps

cos (θ) tan (θ) + cot (θ) = csc (θ)

Subtract

csc (θ)

from both sides

cos (θ) tan (θ) + cot (θ) - csc (θ) = 0

Express with sin, cos

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

Simplify

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

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

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

cos (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= sin (θ)

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

sin (θ) sin (θ) + cos (θ) - 1

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= sin^{2} (θ) + cos (θ) - 1

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

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

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

sin^{2} (θ) + cos (θ) - 1 = 0

Subtract

cos (θ)

from both sides

sin^{2} (θ) - 1 = - cos (θ)

Square both sides

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

Subtract

(- cos (θ))^{2}

from both sides

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

Factor

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

(sin^{2} (θ) - 1)^{2} - cos^{2} (θ)

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Refine

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

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

Solving each part separately

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

sin^{2} (θ) - 1 + cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2 πn

sin^{2} (θ) - 1 + cos (θ) = 0

Rewrite using trig identities

- 1 + cos (θ) + sin^{2} (θ)

Use the Pythagorean identity:

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

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

= cos (θ) - cos^{2} (θ)

cos (θ) - cos^{2} (θ) = 0

Solve by substitution

cos (θ) - cos^{2} (θ) = 0

Let:

cos (θ) = u

u - u^{2} = 0

u - u^{2} = 0 : u = 0, u = 1

u - u^{2} = 0

Write in the standard form

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

- u^{2} + u = 0

Solve with the quadratic formula

- u^{2} + u = 0

Quadratic Equation Formula:

For

a = - 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 0 = 1

1^{2} - 4 (- 1) \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

u_{1} = \frac{- 1 + 1}{2 ( - 1 )}, u_{2} = \frac{- 1 - 1}{2 ( - 1 )}

u = \frac{- 1 + 1}{2 ( - 1 )} : 0

\frac{- 1 + 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 1 - 1}{2 ( - 1 )} : 1

\frac{- 1 - 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 1 - 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 0, u = 1

Substitute back

u = cos (θ)

cos (θ) = 0, cos (θ) = 1

cos (θ) = 0, cos (θ) = 1

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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 = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 1 : θ = 2 πn

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 = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2 πn

sin^{2} (θ) - 1 - cos (θ) = 0 : θ = π + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

sin^{2} (θ) - 1 - cos (θ) = 0

Rewrite using trig identities

- 1 - cos (θ) + sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - cos (θ) - cos^{2} (θ)

- cos (θ) - cos^{2} (θ) = 0

Solve by substitution

- cos (θ) - cos^{2} (θ) = 0

Let:

cos (θ) = u

- u - u^{2} = 0

- u - u^{2} = 0 : u = - 1, u = 0

- u - u^{2} = 0

Write in the standard form

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

- u^{2} - u = 0

Solve with the quadratic formula

- u^{2} - u = 0

Quadratic Equation Formula:

For

a = - 1, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

(- 1)^{2} - 4 (- 1) \cdot 0 = 1

(- 1)^{2} - 4 (- 1) \cdot 0

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 1}{2 ( - 1 )}, u_{2} = \frac{- ( - 1 ) - 1}{2 ( - 1 )}

u = \frac{- ( - 1 ) + 1}{2 ( - 1 )} : - 1

\frac{- ( - 1 ) + 1}{2 ( - 1 )}

Remove parentheses:

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

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

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 1}{2 ( - 1 )} : 0

\frac{- ( - 1 ) - 1}{2 ( - 1 )}

Remove parentheses:

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

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

Subtract the numbers:

1 - 1 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - 1, u = 0

Substitute back

u = cos (θ)

cos (θ) = - 1, cos (θ) = 0

cos (θ) = - 1, cos (θ) = 0

cos (θ) = - 1 : θ = π + 2 πn

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) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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 = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = π + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = 2 πn, θ = π + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

cos (θ) tan (θ) + cot (θ) = csc (θ)

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{2} + 2 πn :

False

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

cos (θ) tan (θ) + cot (θ) = csc (θ)

plug in

θ = \frac{π}{2} + 2 π 1

cos (\frac{π}{2} + 2 π 1) tan (\frac{π}{2} + 2 π 1) + cot (\frac{π}{2} + 2 π 1) = csc (\frac{π}{2} + 2 π 1)

Undefined

\Rightarrow False

Check the solution

\frac{3 π}{2} + 2 πn :

False

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

cos (θ) tan (θ) + cot (θ) = csc (θ)

plug in

θ = \frac{3 π}{2} + 2 π 1

cos (\frac{3 π}{2} + 2 π 1) tan (\frac{3 π}{2} + 2 π 1) + cot (\frac{3 π}{2} + 2 π 1) = csc (\frac{3 π}{2} + 2 π 1)

Undefined

\Rightarrow False

Check the solution

2 πn :

True

2 πn

Plug in

n = 1

2 π 1

For

cos (θ) tan (θ) + cot (θ) = csc (θ)

plug in

θ = 2 π 1

cos (2 π 1) tan (2 π 1) + cot (2 π 1) = csc (2 π 1)

Refine

- \infty = - \infty

\Rightarrow True

Check the solution

π + 2 πn :

False

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

cos (θ) tan (θ) + cot (θ) = csc (θ)

plug in

θ = π + 2 π 1

cos (π + 2 π 1) tan (π + 2 π 1) + cot (π + 2 π 1) = csc (π + 2 π 1)

Refine

- \infty = \infty

\Rightarrow False

θ = 2 πn

Since the equation is undefined for:

2 πn

No Solution for θ \in R

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Frequently Asked Questions (FAQ)

What is the general solution for cos(θ)tan(θ)+cot(θ)=csc(θ) ?
The general solution for cos(θ)tan(θ)+cot(θ)=csc(θ) is No Solution for θ\in\mathbb{R}