What is the general solution for (1-tan^2(θ))/(1+tan^2(θ))=2cos^2(-1) ?

The general solution for (1-tan^2(θ))/(1+tan^2(θ))=2cos^2(-1) is θ=0.47366…+pin,θ=-0.47366…+pin

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(1-tan^2(θ))/(1+tan^2(θ))=2cos^2(-1)

Solution

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

Solution

θ = 0.47366 \dots + πn, θ = - 0.47366 \dots + πn

Degrees

θ = 27.13899 \dots^{\circ} + 18 0^{\circ} n, θ = - 27.13899 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

cos (- 1) = cos (1)

cos (- 1)

Use the following property:

cos (- x) = cos (x)

cos (- 1) = cos (1)

= cos (1)

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

Solve by substitution

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

Let:

tan (θ) = u

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

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

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

Multiply both sides by

1 + u^{2}

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

Multiply both sides by

1 + u^{2}

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

Simplify

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

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

Solve

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

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

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

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

Move

2 cos^{2} (1) (1 + u^{2})

to the left side

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

Subtract

2 cos^{2} (1) (1 + u^{2})

from both sides

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

Simplify

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

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

Expand

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

- 2 cos^{2} (1) (1 + u^{2})

Apply the distributive law:

a (b + c) = ab + a c

a = - 2 cos^{2} (1), b = 1, c = u^{2}

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

Apply minus-plus rules

+ (- a) = - a

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

Multiply the numbers:

2 \cdot 1 = 2

= - 2 cos^{2} (1) - 2 cos^{2} (1) u^{2}

- u^{2} - 2 cos^{2} (1) - 2 cos^{2} (1) u^{2} = - 1

Move

2 cos^{2} (1)

to the right side

- u^{2} - 2 cos^{2} (1) - 2 cos^{2} (1) u^{2} = - 1

Add

2 cos^{2} (1)

to both sides

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

Simplify

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

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

Factor

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

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

Factor out common term

u^{2}

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

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

Divide both sides by

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

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

Divide both sides by

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

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

Simplify

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

Simplify

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

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

Apply the fraction rule:

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

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

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

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

Refine

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

Cancel the common factor:

1 + 2 cos^{2} (1)

= - (- u^{2})

Apply rule

- (- a) = a

= u^{2}

Simplify

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

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

Apply rule

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

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

Apply the fraction rule:

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

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

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

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

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

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

For

x^{2} = f (a)

the solutions are

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

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

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

Substitute back

u = tan (θ)

tan (θ) = \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}, tan (θ) = - \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}

tan (θ) = \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}, tan (θ) = - \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}

tan (θ) = \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1} : θ = arctan (\frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn

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

Apply trig inverse properties

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

General solutions for

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

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (\frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn

θ = arctan (\frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn

tan (θ) = - \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1} : θ = arctan (- \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn

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

Apply trig inverse properties

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

General solutions for

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

tan (x) = - a \Rightarrow x = arctan (- a) + πn

θ = arctan (- \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn

θ = arctan (- \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn

Combine all the solutions

θ = arctan (\frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn, θ = arctan (- \frac{- 2 cos ^{2} ( 1 ) + 1}{2 cos ^{2} ( 1 ) + 1}) + πn

Show solutions in decimal form

θ = 0.47366 \dots + πn, θ = - 0.47366 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (1-tan^2(θ))/(1+tan^2(θ))=2cos^2(-1) ?
The general solution for (1-tan^2(θ))/(1+tan^2(θ))=2cos^2(-1) is θ=0.47366…+pin,θ=-0.47366…+pin