What is the general solution for 1+tan^2(x)=cot^2(x) ?

The general solution for 1+tan^2(x)=cot^2(x) is x=0.66623…+pin,x=2.47535…+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = 0.66623 \dots + πn, x = 2.47535 \dots + πn

Degrees

x = 38.17270 \dots^{\circ} + 18 0^{\circ} n, x = 141.82729 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

cot^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

= 1 - cot^{2} (x) + (\frac{1}{cot ( x )})^{2}

(\frac{1}{cot ( x )})^{2} = \frac{1}{cot ^{2} ( x )}

(\frac{1}{cot ( x )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cot ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cot ^{2} ( x )}

= 1 - cot^{2} (x) + \frac{1}{cot ^{2} ( x )}

1 - cot^{2} (x) + \frac{1}{cot ^{2} ( x )} = 0

Solve by substitution

1 - cot^{2} (x) + \frac{1}{cot ^{2} ( x )} = 0

Let:

cot (x) = u

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

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

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

Multiply both sides by

u^{2}

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

Multiply both sides by

u^{2}

1 \cdot u^{2} - u^{2} u^{2} + \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

Simplify

1 \cdot u^{2} - u^{2} u^{2} + \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

Simplify

1 \cdot u^{2} : u^{2}

1 \cdot u^{2}

Multiply:

1 \cdot u^{2} = u^{2}

= u^{2}

Simplify

- u^{2} u^{2} : - u^{4}

- u^{2} u^{2}

Apply exponent rule:

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

u^{2} u^{2} = u^{2 + 2}

= - u^{2 + 2}

Add the numbers:

2 + 2 = 4

= - u^{4}

Simplify

\frac{1}{u ^{2}} u^{2} : 1

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

Multiply fractions:

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

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

Cancel the common factor:

u^{2}

= 1

Simplify

0 \cdot u^{2} : 0

0 \cdot u^{2}

Apply rule

0 \cdot a = 0

= 0

u^{2} - u^{4} + 1 = 0

u^{2} - u^{4} + 1 = 0

u^{2} - u^{4} + 1 = 0

Solve

u^{2} - u^{4} + 1 = 0 : u = i \frac{- 1 + 5}{2}, u = - i \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}

u^{2} - u^{4} + 1 = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{4} + u^{2} + 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

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

Solve

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

v_{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

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

Separate the solutions

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

v = \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}

v = \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:

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

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

Substitute back

v = u^{2},

solve for

u

Solve

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

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

For

x^{2} = f (a)

the solutions are

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

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

Simplify

- \frac{- 1 + 5}{2} : i \frac{- 1 + 5}{2}

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

Apply radical rule:

- a = - 1 a

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

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

Apply imaginary number rule:

- 1 = i

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

Simplify

- - \frac{- 1 + 5}{2} : - i \frac{- 1 + 5}{2}

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

Simplify

- \frac{- 1 + 5}{2} : i \frac{5 - 1}{2}

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

Apply radical rule:

- a = - 1 a

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

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

Apply imaginary number rule:

- 1 = i

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

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

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

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

Solve

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

u^{2} = \frac{1 + 5}{2}

For

x^{2} = f (a)

the solutions are

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

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

The solutions are

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

Substitute back

u = cot (x)

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

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

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

No Solution

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

No Solution

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

No Solution

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

No Solution

cot (x) = \frac{1 + 5}{2} : x = arccot \frac{1 + 5}{2} + πn

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

Apply trig inverse properties

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

General solutions for

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

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot \frac{1 + 5}{2} + πn

x = arccot \frac{1 + 5}{2} + πn

cot (x) = - \frac{1 + 5}{2} : x = arccot - \frac{1 + 5}{2} + πn

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

Apply trig inverse properties

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

General solutions for

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

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot - \frac{1 + 5}{2} + πn

x = arccot - \frac{1 + 5}{2} + πn

Combine all the solutions

x = arccot \frac{1 + 5}{2} + πn, x = arccot - \frac{1 + 5}{2} + πn

Show solutions in decimal form

x = 0.66623 \dots + πn, x = 2.47535 \dots + πn

Popular Examples

1-4tan(3x+1)=17

1 - 4 tan (3 x + 1) = 17

sin(x)cos(x)+sin(x)=3cos^2(x)+3cos(x)

sin (x) cos (x) + sin (x) = 3 cos^{2} (x) + 3 cos (x)

2-2sin^2(x)=-3cos(x)+2

2 - 2 sin^{2} (x) = - 3 cos (x) + 2

4tan(θ)+7=0

4 tan (θ) + 7 = 0

arctan(x)+arctan(3x)+arctan(7x)=180

arctan (x) + arctan (3 x) + arctan (7 x) = 18 0^{\circ}

Frequently Asked Questions (FAQ)

What is the general solution for 1+tan^2(x)=cot^2(x) ?
The general solution for 1+tan^2(x)=cot^2(x) is x=0.66623…+pin,x=2.47535…+pin