What is the general solution for (tan(x))/(sec(x))=cot(x) ?

The general solution for (tan(x))/(sec(x))=cot(x) is x=0.90455…+2pin,x=2pi-0.90455…+2pin

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Popular Trigonometry >

(tan(x))/(sec(x))=cot(x)

Solution

\frac{tan ( x )}{sec ( x )} = cot (x)

Solution

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

Degrees

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

Solution steps

\frac{tan ( x )}{sec ( x )} = cot (x)

Subtract

cot (x)

from both sides

\frac{tan ( x )}{sec ( x )} - cot (x) = 0

Simplify

\frac{tan ( x )}{sec ( x )} - cot (x) : \frac{tan ( x ) - cot ( x ) sec ( x )}{sec ( x )}

\frac{tan ( x )}{sec ( x )} - cot (x)

Convert element to fraction:

cot (x) = \frac{cot ( x ) sec ( x )}{sec ( x )}

= \frac{tan ( x )}{sec ( x )} - \frac{cot ( x ) sec ( x )}{sec ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{tan ( x ) - cot ( x ) sec ( x )}{sec ( x )}

\frac{tan ( x ) - cot ( x ) sec ( x )}{sec ( x )} = 0

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

tan (x) - cot (x) sec (x) = 0

Express with sin, cos

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

Simplify

\frac{sin ( x )}{cos ( x )} - \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{cos ( x )} : \frac{sin ^{2} ( x ) - cos ( x )}{cos ( x ) sin ( x )}

\frac{sin ( x )}{cos ( x )} - \frac{cos ( x )}{sin ( x )} \cdot \frac{1}{cos ( x )}

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

= \frac{sin ( x )}{cos ( x )} - \frac{1}{sin ( x )}

Least Common Multiplier of

cos (x), sin (x) : cos (x) sin (x)

cos (x), sin (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (x)

sin (x)

= cos (x) sin (x)

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

cos (x) sin (x)

For

\frac{sin ( x )}{cos ( x )} :

multiply the denominator and numerator by

sin (x)

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

For

\frac{1}{sin ( x )} :

multiply the denominator and numerator by

cos (x)

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

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

Since the denominators are equal, combine the fractions:

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

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

\frac{sin ^{2} ( x ) - cos ( x )}{cos ( x ) sin ( x )} = 0

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

sin^{2} (x) - cos (x) = 0

Add

cos (x)

to both sides

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

Square both sides

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

Subtract

cos^{2} (x)

from both sides

sin^{4} (x) - cos^{2} (x) = 0

Factor

sin^{4} (x) - cos^{2} (x) : (sin^{2} (x) + cos (x)) (sin^{2} (x) - cos (x))

sin^{4} (x) - cos^{2} (x)

Apply exponent rule:

a^{b c} = (a^{b})^{c}

sin^{4} (x) = (sin^{2} (x))^{2}

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

Apply Difference of Two Squares Formula:

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

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

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

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

Solving each part separately

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

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

sin^{2} (x) + cos (x) = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = 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 ( - 1 ) \cdot 1}{2 ( - 1 )}

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 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 ( - 1 )}

Separate the solutions

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

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

\frac{- 1 + 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

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

\frac{- 1 - 5}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

- 1 - 5 = - (1 + 5)

= \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 (x)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

cos (x) = \frac{1 + 5}{2} :

No Solution

cos (x) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

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

sin^{2} (x) - cos (x) = 0

Rewrite using trig identities

- cos (x) + sin^{2} (x)

Use the Pythagorean identity:

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

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

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

1 - cos (x) - cos^{2} (x) = 0

Solve by substitution

1 - cos (x) - cos^{2} (x) = 0

Let:

cos (x) = u

1 - u - u^{2} = 0

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

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

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

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

Apply rule

- (- a) = a

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

(- 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 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

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

Separate the solutions

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

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

\frac{- ( - 1 ) + 5}{2 ( - 1 )}

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

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

u = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : \frac{5 - 1}{2}

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

Remove parentheses:

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

Apply the fraction rule:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = cos (x)

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

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

cos (x) = - \frac{1 + 5}{2} :

No Solution

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

- 1 \leq cos (x) \leq 1

No Solution

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

cos (x) = \frac{5 - 1}{2}

Apply trig inverse properties

cos (x) = \frac{5 - 1}{2}

General solutions for

cos (x) = \frac{5 - 1}{2}

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

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

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

Combine all the solutions

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

Combine all the solutions

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

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

\frac{tan ( x )}{sec ( x )} = cot (x)

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

Check the solution

arccos (- \frac{- 1 + 5}{2}) + 2 πn :

False

arccos (- \frac{- 1 + 5}{2}) + 2 πn

Plug in

n = 1

arccos (- \frac{- 1 + 5}{2}) + 2 π 1

For

\frac{tan ( x )}{sec ( x )} = cot (x)

plug in

x = arccos (- \frac{- 1 + 5}{2}) + 2 π 1

\frac{tan ( arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )}{sec ( arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )} = cot (arccos (- \frac{- 1 + 5}{2}) + 2 π 1)

Refine

0.78615 \dots = - 0.78615 \dots

\Rightarrow False

Check the solution

- arccos (- \frac{- 1 + 5}{2}) + 2 πn :

False

- arccos (- \frac{- 1 + 5}{2}) + 2 πn

Plug in

n = 1

- arccos (- \frac{- 1 + 5}{2}) + 2 π 1

For

\frac{tan ( x )}{sec ( x )} = cot (x)

plug in

x = - arccos (- \frac{- 1 + 5}{2}) + 2 π 1

\frac{tan ( - arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )}{sec ( - arccos ( - \frac{- 1 + 5}{2} ) + 2 π 1 )} = cot (- arccos (- \frac{- 1 + 5}{2}) + 2 π 1)

Refine

- 0.78615 \dots = 0.78615 \dots

\Rightarrow False

Check the solution

arccos (\frac{5 - 1}{2}) + 2 πn :

True

arccos (\frac{5 - 1}{2}) + 2 πn

Plug in

n = 1

arccos (\frac{5 - 1}{2}) + 2 π 1

For

\frac{tan ( x )}{sec ( x )} = cot (x)

plug in

x = arccos (\frac{5 - 1}{2}) + 2 π 1

\frac{tan ( arccos ( \frac{5 - 1}{2} ) + 2 π 1 )}{sec ( arccos ( \frac{5 - 1}{2} ) + 2 π 1 )} = cot (arccos (\frac{5 - 1}{2}) + 2 π 1)

Refine

0.78615 \dots = 0.78615 \dots

\Rightarrow True

Check the solution

2 π - arccos (\frac{5 - 1}{2}) + 2 πn :

True

2 π - arccos (\frac{5 - 1}{2}) + 2 πn

Plug in

n = 1

2 π - arccos (\frac{5 - 1}{2}) + 2 π 1

For

\frac{tan ( x )}{sec ( x )} = cot (x)

plug in

x = 2 π - arccos (\frac{5 - 1}{2}) + 2 π 1

\frac{tan ( 2 π - arccos ( \frac{5 - 1}{2} ) + 2 π 1 )}{sec ( 2 π - arccos ( \frac{5 - 1}{2} ) + 2 π 1 )} = cot (2 π - arccos (\frac{5 - 1}{2}) + 2 π 1)

Refine

- 0.78615 \dots = - 0.78615 \dots

\Rightarrow True

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

Show solutions in decimal form

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (tan(x))/(sec(x))=cot(x) ?
The general solution for (tan(x))/(sec(x))=cot(x) is x=0.90455…+2pin,x=2pi-0.90455…+2pin