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Popular Trigonometry >

solvefor t,x=arccos(1/(sqrt(1+t^2)))

Solution

solve for

t, x = arccos (\frac{1}{1 + t ^{2}})

Solution

t = \frac{1 - cos ^{2} ( x )}{cos ( x )}, t = - \frac{1 - cos ^{2} ( x )}{cos ( x )}

Solution steps

x = arccos (\frac{1}{1 + t ^{2}})

Switch sides

arccos (\frac{1}{1 + t ^{2}}) = x

Apply trig inverse properties

arccos (\frac{1}{1 + t ^{2}}) = x

arccos (x) = a \Rightarrow x = cos (a)

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

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

Solve

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

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

Multiply both sides by

1 + t^{2}

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

Simplify

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

Switch sides

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

Divide both sides by

cos (x)

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

Divide both sides by

cos (x)

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

Simplify

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

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

Square both sides:

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

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

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

Expand

(1 + t^{2})^{2} : 1 + t^{2}

(1 + t^{2})^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= ((1 + t^{2})^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= (1 + t^{2})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 1 + t^{2}

Expand

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Solve

1 + t^{2} = \frac{1}{cos ^{2} ( x )} : t = \frac{1 - cos ^{2} ( x )}{cos ( x )}, t = - \frac{1 - cos ^{2} ( x )}{cos ( x )}

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

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

t = \frac{1}{cos ^{2} ( x )} - 1, t = - \frac{1}{cos ^{2} ( x )} - 1

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

Apply radical rule:

assuming

a \geq 0, b \geq 0

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

Apply radical rule:

assuming

a \geq 0

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

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

Simplify

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

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

Apply radical rule:

assuming

a \geq 0, b \geq 0

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

Apply radical rule:

assuming

a \geq 0

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

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

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

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

t = \frac{1 - cos ^{2} ( x )}{cos ( x )}, t = - \frac{1 - cos ^{2} ( x )}{cos ( x )}

t = \frac{1 - cos ^{2} ( x )}{cos ( x )}, t = - \frac{1 - cos ^{2} ( x )}{cos ( x )}

Verify Solutions:

t = \frac{1 - cos ^{2} ( x )}{cos ( x )} {x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn}, t = - \frac{1 - cos ^{2} ( x )}{cos ( x )} {x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn}

Check the solutions by plugging them into

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

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

Plug

t = \frac{1 - cos ^{2} ( x )}{cos ( x )} : \frac{1}{1 + ( \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x) \Rightarrow x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

\frac{1}{1 + ( \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x)

Solve by substitution

\frac{1}{1 + ( \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x)

Let:

cos (x) = u

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

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

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

Multiply both sides by

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

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

Simplify

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

Square both sides:

1 = 1

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

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

Expand

1^{2} : 1

1^{2}

Apply rule

1^{a} = 1

= 1

Expand

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

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

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

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

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

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

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

Expand

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

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

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

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

Apply exponent rule:

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

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

(1 - u^{2})^{2} : 1 - u^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

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

= (1 - u^{2})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 1 - u^{2}

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

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

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

Apply the distributive law:

a (b + c) = ab + a c

a = u^{2}, b = 1, c = \frac{1 - u ^{2}}{u ^{2}}

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

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

Simplify

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

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

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

1 \cdot u^{2}

Multiply:

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

= u^{2}

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

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

Multiply fractions:

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

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

Cancel the common factor:

u^{2}

= 1 - u^{2}

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

Group like terms

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

Add similar elements:

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

= 1

= 1

= 1

1 = 1

1 = 1

Both sides are equal

True for all u

Verify Solutions:

u < - 1

False

, u = - 1

False

, - 1 < u < 1

False

, u = 1

True

, u > 1

False

Check the solutions by plugging them into

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

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

Plug

u < - 1 : \frac{1}{1 + ( \frac{1 - u ^{2}}{u} ) ^{2}} \neq = u \Rightarrow

False

The solution is

u = 1

Substitute back

u = cos (x)

cos (x) = 1

cos (x) = 1

cos (x) = u = 1 : x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

cos (x) = u = 1

Apply trig inverse properties

cos (x) = u = 1

General solutions for

cos (x) = u = 1

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

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

Combine all the solutions

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

Plug

t = - \frac{1 - cos ^{2} ( x )}{cos ( x )} : \frac{1}{1 + ( - \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x) \Rightarrow x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

\frac{1}{1 + ( - \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x)

Solve by substitution

\frac{1}{1 + ( - \frac{1 - c o s ^{2} ( x )}{c o s ( x )} ) ^{2}} = cos (x)

Let:

cos (x) = u

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

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

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

Multiply both sides by

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

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

Simplify

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

Expand

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

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

Combine same powers :

\frac{x}{y} = \frac{x}{y}

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

Expand

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

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

Apply rule

1 = 1

= 1

= 1

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

Square both sides:

1 = 1

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

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

Expand

1^{2} : 1

1^{2}

Apply rule

1^{a} = 1

= 1

Expand

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

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

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

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

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

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

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

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

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

Expand

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

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

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

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

Apply exponent rule:

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

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

(1 - u^{2})^{2} : 1 - u^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

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

= (1 - u^{2})^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 1 - u^{2}

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

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

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

Apply the distributive law:

a (b + c) = ab + a c

a = u^{2}, b = 1, c = \frac{1 - u ^{2}}{u ^{2}}

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

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

Simplify

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

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

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

1 \cdot u^{2}

Multiply:

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

= u^{2}

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

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

Multiply fractions:

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

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

Cancel the common factor:

u^{2}

= 1 - u^{2}

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

Group like terms

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

Add similar elements:

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

= 1

= 1

= 1

1 = 1

1 = 1

Both sides are equal

True for all u

Verify Solutions:

u < - 1

False

, u = - 1

False

, - 1 < u < 1

False

, u = 1

True

, u > 1

False

Check the solutions by plugging them into

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

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

Plug

u < - 1 : \frac{1}{1 + ( - \frac{1 - u ^{2}}{u} ) ^{2}} \neq = u \Rightarrow

False

The solution is

u = 1

Substitute back

u = cos (x)

cos (x) = 1

cos (x) = 1

cos (x) = u = 1 : x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

cos (x) = u = 1

Apply trig inverse properties

cos (x) = u = 1

General solutions for

cos (x) = u = 1

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

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

Combine all the solutions

x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn

The solutions are

t = \frac{1 - cos ^{2} ( x )}{cos ( x )} {x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn}, t = - \frac{1 - cos ^{2} ( x )}{cos ( x )} {x = arccos (u = 1) + 2 πn, x = - arccos (u = 1) + 2 πn}

t = \frac{1 - cos ^{2} ( x )}{cos ( x )}, t = - \frac{1 - cos ^{2} ( x )}{cos ( x )}

Graph

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