What is the general solution for tan^2(x)=cot^2(x) ?

The general solution for tan^2(x)=cot^2(x) is x= pi/4+pin,x=(3pi)/4+pin

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Popular Trigonometry >

tan^2(x)=cot^2(x)

Solution

tan^{2} (x) = cot^{2} (x)

Solution

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) = cot^{2} (x)

Subtract

cot^{2} (x)

from both sides

tan^{2} (x) - cot^{2} (x) = 0

Factor

tan^{2} (x) - cot^{2} (x) : (tan (x) + cot (x)) (tan (x) - cot (x))

tan^{2} (x) - cot^{2} (x)

Apply Difference of Two Squares Formula:

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

tan^{2} (x) - cot^{2} (x) = (tan (x) + cot (x)) (tan (x) - cot (x))

= (tan (x) + cot (x)) (tan (x) - cot (x))

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

Solving each part separately

tan (x) + cot (x) = 0 or tan (x) - cot (x) = 0

tan (x) + cot (x) = 0 :

No Solution

tan (x) + cot (x) = 0

Rewrite using trig identities

cot (x) + tan (x)

Use the basic trigonometric identity:

tan (x) = \frac{1}{cot ( x )}

= cot (x) + \frac{1}{cot ( x )}

cot (x) + \frac{1}{cot ( x )} = 0

Solve by substitution

cot (x) + \frac{1}{cot ( x )} = 0

Let:

cot (x) = u

u + \frac{1}{u} = 0

u + \frac{1}{u} = 0 : u = i, u = - i

u + \frac{1}{u} = 0

Multiply both sides by

u

u + \frac{1}{u} = 0

Multiply both sides by

u

uu + \frac{1}{u} u = 0 \cdot u

Simplify

uu + \frac{1}{u} u = 0 \cdot u

Simplify

uu : u^{2}

uu

Apply exponent rule:

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

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot u}{u}

Cancel the common factor:

u

= 1

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

u^{2} + 1 = 0

u^{2} + 1 = 0

u^{2} + 1 = 0

Solve

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

u^{2} + 1 = 0

Move

1

to the right side

u^{2} + 1 = 0

Subtract

1

from both sides

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

Simplify

u^{2} = - 1

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

u = i, u = - i

Substitute back

u = cot (x)

cot (x) = i, cot (x) = - i

cot (x) = i, cot (x) = - i

cot (x) = i :

No Solution

cot (x) = i

No Solution

cot (x) = - i :

No Solution

cot (x) = - i

No Solution

Combine all the solutions

No Solution

tan (x) - cot (x) = 0 : x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

tan (x) - cot (x) = 0

Rewrite using trig identities

- cot (x) + tan (x)

Use the basic trigonometric identity:

tan (x) = \frac{1}{cot ( x )}

= - cot (x) + \frac{1}{cot ( x )}

- cot (x) + \frac{1}{cot ( x )} = 0

Solve by substitution

- cot (x) + \frac{1}{cot ( x )} = 0

Let:

cot (x) = u

- u + \frac{1}{u} = 0

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

- u + \frac{1}{u} = 0

Multiply both sides by

u

- u + \frac{1}{u} = 0

Multiply both sides by

u

- uu + \frac{1}{u} u = 0 \cdot u

Simplify

- uu + \frac{1}{u} u = 0 \cdot u

Simplify

- uu : - u^{2}

- uu

Apply exponent rule:

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

uu = u^{1 + 1}

= - u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - u^{2}

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot u}{u}

Cancel the common factor:

u

= 1

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

- u^{2} + 1 = 0

- u^{2} + 1 = 0

- u^{2} + 1 = 0

Solve

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

- u^{2} + 1 = 0

Move

1

to the right side

- u^{2} + 1 = 0

Subtract

1

from both sides

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

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 1}

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

u = 1, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- u + \frac{1}{u}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1, u = - 1

Substitute back

u = cot (x)

cot (x) = 1, cot (x) = - 1

cot (x) = 1, cot (x) = - 1

cot (x) = 1 : x = \frac{π}{4} + πn

cot (x) = 1

General solutions for

cot (x) = 1

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

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

cot (x) = - 1 : x = \frac{3 π}{4} + πn

cot (x) = - 1

General solutions for

cot (x) = - 1

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

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)=cot^2(x) ?
The general solution for tan^2(x)=cot^2(x) is x= pi/4+pin,x=(3pi)/4+pin