What is the general solution for (tan(θ)cot(θ))/(sec^2(θ))=cot(θ) ?

The general solution for (tan(θ)cot(θ))/(sec^2(θ))=cot(θ) is No Solution for θ\in\mathbb{R}

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Popular Trigonometry >

(tan(θ)cot(θ))/(sec^2(θ))=cot(θ)

Solution

\frac{tan ( θ ) cot ( θ )}{sec ^{2} ( θ )} = cot (θ)

Solution

No Solution for θ \in R

Solution steps

\frac{tan ( θ ) cot ( θ )}{sec ^{2} ( θ )} = cot (θ)

Subtract

cot (θ)

from both sides

\frac{tan ( θ ) cot ( θ )}{sec ^{2} ( θ )} - cot (θ) = 0

Simplify

\frac{tan ( θ ) cot ( θ )}{sec ^{2} ( θ )} - cot (θ) : \frac{tan ( θ ) cot ( θ ) - sec ^{2} ( θ ) cot ( θ )}{sec ^{2} ( θ )}

\frac{tan ( θ ) cot ( θ )}{sec ^{2} ( θ )} - cot (θ)

Convert element to fraction:

cot (θ) = \frac{cot ( θ ) sec ^{2} ( θ )}{sec ^{2} ( θ )}

= \frac{tan ( θ ) cot ( θ )}{sec ^{2} ( θ )} - \frac{cot ( θ ) sec ^{2} ( θ )}{sec ^{2} ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{tan ( θ ) cot ( θ ) - cot ( θ ) sec ^{2} ( θ )}{sec ^{2} ( θ )}

\frac{tan ( θ ) cot ( θ ) - sec ^{2} ( θ ) cot ( θ )}{sec ^{2} ( θ )} = 0

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

tan (θ) cot (θ) - sec^{2} (θ) cot (θ) = 0

Factor

tan (θ) cot (θ) - sec^{2} (θ) cot (θ) : cot (θ) (tan (θ) - sec^{2} (θ))

tan (θ) cot (θ) - sec^{2} (θ) cot (θ)

Factor out common term

cot (θ)

= cot (θ) (tan (θ) - sec^{2} (θ))

cot (θ) (tan (θ) - sec^{2} (θ)) = 0

Solving each part separately

cot (θ) = 0 or tan (θ) - sec^{2} (θ) = 0

cot (θ) = 0 : θ = \frac{π}{2} + πn

cot (θ) = 0

General solutions for

cot (θ) = 0

cot (x)

periodicity table with

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

tan (θ) - sec^{2} (θ) = 0 :

No Solution

tan (θ) - sec^{2} (θ) = 0

Rewrite using trig identities

- sec^{2} (θ) + tan (θ)

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

= - (tan^{2} (θ) + 1) + tan (θ)

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

- (tan^{2} (θ) + 1)

Distribute parentheses

= - (tan^{2} (θ)) - (1)

Apply minus-plus rules

+ (- a) = - a

= - tan^{2} (θ) - 1

= - tan^{2} (θ) - 1 + tan (θ)

- 1 + tan (θ) - tan^{2} (θ) = 0

Solve by substitution

- 1 + tan (θ) - tan^{2} (θ) = 0

Let:

tan (θ) = u

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

- 1 + u - u^{2} = 0 : u = \frac{1}{2} - i \frac{3}{2}, u = \frac{1}{2} + i \frac{3}{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 ( - 1 ) ( - 1 )}{2 ( - 1 )}

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

Simplify

1^{2} - 4 (- 1) (- 1) : 3 i

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) (- 1)

Apply rule

- (- a) = a

= 1 - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 - 4

Subtract the numbers:

1 - 4 = - 3

= - 3

Apply radical rule:

- a = - 1 a

- 3 = - 1 3

= - 1 3

Apply imaginary number rule:

- 1 = i

= 3 i

u_{1, 2} = \frac{- 1 \pm 3 i}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 3 i}{2 ( - 1 )}, u_{2} = \frac{- 1 - 3 i}{2 ( - 1 )}

u = \frac{- 1 + 3 i}{2 ( - 1 )} : \frac{1}{2} - i \frac{3}{2}

\frac{- 1 + 3 i}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 3 i}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 3 i}{- 2}

Apply the fraction rule:

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

= - \frac{- 1 + 3 i}{2}

Rewrite

- \frac{- 1 + 3 i}{2}

in standard complex form:

\frac{1}{2} - \frac{3}{2} i

- \frac{- 1 + 3 i}{2}

Apply the fraction rule:

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

\frac{- 1 + 3 i}{2} = - (- \frac{1}{2}) - (\frac{3 i}{2})

= - (- \frac{1}{2}) - (\frac{3 i}{2})

Remove parentheses:

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

= \frac{1}{2} - \frac{3 i}{2}

= \frac{1}{2} - \frac{3}{2} i

u = \frac{- 1 - 3 i}{2 ( - 1 )} : \frac{1}{2} + i \frac{3}{2}

\frac{- 1 - 3 i}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 3 i}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 3 i}{- 2}

Apply the fraction rule:

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

= - \frac{- 1 - 3 i}{2}

Rewrite

- \frac{- 1 - 3 i}{2}

in standard complex form:

\frac{1}{2} + \frac{3}{2} i

- \frac{- 1 - 3 i}{2}

Apply the fraction rule:

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

\frac{- 1 - 3 i}{2} = - (- \frac{1}{2}) - (- \frac{3 i}{2})

= - (- \frac{1}{2}) - (- \frac{3 i}{2})

Apply rule

- (- a) = a

= \frac{1}{2} + \frac{3 i}{2}

= \frac{1}{2} + \frac{3}{2} i

The solutions to the quadratic equation are:

u = \frac{1}{2} - i \frac{3}{2}, u = \frac{1}{2} + i \frac{3}{2}

Substitute back

u = tan (θ)

tan (θ) = \frac{1}{2} - i \frac{3}{2}, tan (θ) = \frac{1}{2} + i \frac{3}{2}

tan (θ) = \frac{1}{2} - i \frac{3}{2}, tan (θ) = \frac{1}{2} + i \frac{3}{2}

tan (θ) = \frac{1}{2} - i \frac{3}{2} :

No Solution

tan (θ) = \frac{1}{2} - i \frac{3}{2}

No Solution

tan (θ) = \frac{1}{2} + i \frac{3}{2} :

No Solution

tan (θ) = \frac{1}{2} + i \frac{3}{2}

No Solution

Combine all the solutions

No Solution

Combine all the solutions

θ = \frac{π}{2} + πn

Since the equation is undefined for:

\frac{π}{2} + πn

No Solution for θ \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (tan(θ)cot(θ))/(sec^2(θ))=cot(θ) ?
The general solution for (tan(θ)cot(θ))/(sec^2(θ))=cot(θ) is No Solution for θ\in\mathbb{R}