What is the general solution for tan(2t)+tan(t)=1-tan(2t)tan(t) ?

The general solution for tan(2t)+tan(t)=1-tan(2t)tan(t) is t=1.30899…+pin,t=0.26179…+pin

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tan(2t)+tan(t)=1-tan(2t)tan(t)

Solution

tan (2 t) + tan (t) = 1 - tan (2 t) tan (t)

Solution

t = 1.30899 \dots + πn, t = 0.26179 \dots + πn

Degrees

t = 7 5^{\circ} + 18 0^{\circ} n, t = 1 5^{\circ} + 18 0^{\circ} n

Solution steps

tan (2 t) + tan (t) = 1 - tan (2 t) tan (t)

Subtract

1 - tan (2 t) tan (t)

from both sides

tan (2 t) + tan (t) - 1 + tan (2 t) tan (t) = 0

Rewrite using trig identities

- 1 + tan (2 t) + tan (t) + tan (2 t) tan (t)

Use the Double Angle identity:

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

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

Simplify

- 1 + \frac{2 tan ( t )}{1 - tan ^{2} ( t )} + tan (t) + \frac{2 tan ( t )}{1 - tan ^{2} ( t )} tan (t) : \frac{- 3 tan ( t ) + tan ^{2} ( t )}{tan ( t ) - 1} - 1

- 1 + \frac{2 tan ( t )}{1 - tan ^{2} ( t )} + tan (t) + \frac{2 tan ( t )}{1 - tan ^{2} ( t )} tan (t)

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

\frac{2 tan ( t )}{1 - tan ^{2} ( t )} tan (t)

Multiply fractions:

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

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

2 tan (t) tan (t) = 2 tan^{2} (t)

2 tan (t) tan (t)

Apply exponent rule:

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

tan (t) tan (t) = tan^{1 + 1} (t)

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

Add the numbers:

1 + 1 = 2

= 2 tan^{2} (t)

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

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

Combine the fractions

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

Apply rule

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

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

Factor out common term

2 tan (t) : 2 tan (t) (tan (t) + 1)

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

Apply exponent rule:

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

tan^{2} (t) = tan (t) tan (t)

= 2 tan (t) tan (t) + 2 tan (t)

Factor out common term

2 tan (t)

= 2 tan (t) (tan (t) + 1)

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

Factor

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

1 - tan^{2} (t)

Factor out common term

- 1

= - (tan^{2} (t) - 1)

Factor

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

tan^{2} (t) - 1

Rewrite

1

1^{2}

= tan^{2} (t) - 1^{2}

Apply Difference of Two Squares Formula:

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

tan^{2} (t) - 1^{2} = (tan (t) + 1) (tan (t) - 1)

= (tan (t) + 1) (tan (t) - 1)

= - (tan (t) + 1) (tan (t) - 1)

= - \frac{2 tan ( t ) ( tan ( t ) + 1 )}{( tan ( t ) + 1 ) ( tan ( t ) - 1 )}

Cancel the common factor:

tan (t) + 1

= - \frac{2 tan ( t )}{tan ( t ) - 1}

= - 1 - \frac{2 tan ( t )}{tan ( t ) - 1} + tan (t)

Combine the fractions

- \frac{2 tan ( t )}{tan ( t ) - 1} + tan (t) : \frac{- 3 tan ( t ) + tan ^{2} ( t )}{tan ( t ) - 1}

- \frac{2 tan ( t )}{tan ( t ) - 1} + tan (t)

Convert element to fraction:

tan (t) = \frac{tan ( t ) ( tan ( t ) - 1 )}{tan ( t ) - 1}

= - \frac{2 tan ( t )}{tan ( t ) - 1} + \frac{tan ( t ) ( tan ( t ) - 1 )}{tan ( t ) - 1}

Since the denominators are equal, combine the fractions:

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

= \frac{- 2 tan ( t ) + tan ( t ) ( tan ( t ) - 1 )}{tan ( t ) - 1}

Expand

- 2 tan (t) + tan (t) (tan (t) - 1) : - 3 tan (t) + tan^{2} (t)

- 2 tan (t) + tan (t) (tan (t) - 1)

Expand

tan (t) (tan (t) - 1) : tan^{2} (t) - tan (t)

tan (t) (tan (t) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = tan (t), b = tan (t), c = 1

= tan (t) tan (t) - tan (t) \cdot 1

= tan (t) tan (t) - 1 \cdot tan (t)

Simplify

tan (t) tan (t) - 1 \cdot tan (t) : tan^{2} (t) - tan (t)

tan (t) tan (t) - 1 \cdot tan (t)

tan (t) tan (t) = tan^{2} (t)

tan (t) tan (t)

Apply exponent rule:

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

tan (t) tan (t) = tan^{1 + 1} (t)

= tan^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= tan^{2} (t)

1 \cdot tan (t) = tan (t)

1 \cdot tan (t)

Multiply:

1 \cdot tan (t) = tan (t)

= tan (t)

= tan^{2} (t) - tan (t)

= tan^{2} (t) - tan (t)

= - 2 tan (t) + tan^{2} (t) - tan (t)

Add similar elements:

- 2 tan (t) - tan (t) = - 3 tan (t)

= - 3 tan (t) + tan^{2} (t)

= \frac{- 3 tan ( t ) + tan ^{2} ( t )}{tan ( t ) - 1}

= \frac{tan ^{2} ( t ) - 3 tan ( t )}{tan ( t ) - 1} - 1

= \frac{- 3 tan ( t ) + tan ^{2} ( t )}{tan ( t ) - 1} - 1

- 1 + \frac{tan ^{2} ( t ) - 3 tan ( t )}{- 1 + tan ( t )} = 0

Solve by substitution

- 1 + \frac{tan ^{2} ( t ) - 3 tan ( t )}{- 1 + tan ( t )} = 0

Let:

tan (t) = u

- 1 + \frac{u ^{2} - 3 u}{- 1 + u} = 0

- 1 + \frac{u ^{2} - 3 u}{- 1 + u} = 0 : u = 2 + 3, u = 2 - 3

- 1 + \frac{u ^{2} - 3 u}{- 1 + u} = 0

Multiply both sides by

- 1 + u

- 1 + \frac{u ^{2} - 3 u}{- 1 + u} = 0

Multiply both sides by

- 1 + u

- 1 \cdot (- 1 + u) + \frac{u ^{2} - 3 u}{- 1 + u} (- 1 + u) = 0 \cdot (- 1 + u)

Simplify

- 1 \cdot (- 1 + u) + \frac{u ^{2} - 3 u}{- 1 + u} (- 1 + u) = 0 \cdot (- 1 + u)

Simplify

- 1 \cdot (- 1 + u) : - (- 1 + u)

- 1 \cdot (- 1 + u)

Multiply:

1 \cdot (- 1 + u) = (- 1 + u)

= - (u - 1)

Simplify

\frac{u ^{2} - 3 u}{- 1 + u} (- 1 + u) : u^{2} - 3 u

\frac{u ^{2} - 3 u}{- 1 + u} (- 1 + u)

Multiply fractions:

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

= \frac{( u ^{2} - 3 u ) ( - 1 + u )}{- 1 + u}

Cancel the common factor:

- 1 + u

= u^{2} - 3 u

Simplify

0 \cdot (- 1 + u) : 0

0 \cdot (- 1 + u)

Apply rule

0 \cdot a = 0

= 0

- (- 1 + u) + u^{2} - 3 u = 0

- (- 1 + u) + u^{2} - 3 u = 0

- (- 1 + u) + u^{2} - 3 u = 0

Solve

- (- 1 + u) + u^{2} - 3 u = 0 : u = 2 + 3, u = 2 - 3

- (- 1 + u) + u^{2} - 3 u = 0

Expand

- (- 1 + u) + u^{2} - 3 u : u^{2} - 4 u + 1

- (- 1 + u) + u^{2} - 3 u

- (- 1 + u) : 1 - u

- (- 1 + u)

Distribute parentheses

= - (- 1) - (u)

Apply minus-plus rules

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

= 1 - u

= 1 - u + u^{2} - 3 u

Simplify

1 - u + u^{2} - 3 u : u^{2} - 4 u + 1

1 - u + u^{2} - 3 u

Group like terms

= u^{2} - u - 3 u + 1

Add similar elements:

- u - 3 u = - 4 u

= u^{2} - 4 u + 1

= u^{2} - 4 u + 1

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 1, b = - 4, c = 1

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

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

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

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

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4^{2} - 4

4^{2} = 16

= 16 - 4

Subtract the numbers:

16 - 4 = 12

= 12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

u_{1, 2} = \frac{- ( - 4 ) \pm 2 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 2 3}{2 \cdot 1}, u_{2} = \frac{- ( - 4 ) - 2 3}{2 \cdot 1}

u = \frac{- ( - 4 ) + 2 3}{2 \cdot 1} : 2 + 3

\frac{- ( - 4 ) + 2 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 + 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4 + 2 3}{2}

Factor

4 + 23 : 2 (2 + 3)

4 + 23

Rewrite as

= 2 \cdot 2 + 23

Factor out common term

2

= 2 (2 + 3)

= \frac{2 ( 2 + 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2 + 3

u = \frac{- ( - 4 ) - 2 3}{2 \cdot 1} : 2 - 3

\frac{- ( - 4 ) - 2 3}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{4 - 2 3}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4 - 2 3}{2}

Factor

4 - 23 : 2 (2 - 3)

4 - 23

Rewrite as

= 2 \cdot 2 - 23

Factor out common term

2

= 2 (2 - 3)

= \frac{2 ( 2 - 3 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2 - 3

The solutions to the quadratic equation are:

u = 2 + 3, u = 2 - 3

u = 2 + 3, u = 2 - 3

Verify Solutions

Find undefined (singularity) points:

u = 1

Take the denominator(s) of

- 1 + \frac{u ^{2} - 3 u}{- 1 + u}

and compare to zero

Solve

- 1 + u = 0 : u = 1

- 1 + u = 0

Move

1

to the right side

- 1 + u = 0

Add

1

to both sides

- 1 + u + 1 = 0 + 1

Simplify

u = 1

u = 1

The following points are undefined

u = 1

Combine undefined points with solutions:

u = 2 + 3, u = 2 - 3

Substitute back

u = tan (t)

tan (t) = 2 + 3, tan (t) = 2 - 3

tan (t) = 2 + 3, tan (t) = 2 - 3

tan (t) = 2 + 3 : t = arctan (2 + 3) + πn

tan (t) = 2 + 3

Apply trig inverse properties

tan (t) = 2 + 3

General solutions for

tan (t) = 2 + 3

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

t = arctan (2 + 3) + πn

t = arctan (2 + 3) + πn

tan (t) = 2 - 3 : t = arctan (2 - 3) + πn

tan (t) = 2 - 3

Apply trig inverse properties

tan (t) = 2 - 3

General solutions for

tan (t) = 2 - 3

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

t = arctan (2 - 3) + πn

t = arctan (2 - 3) + πn

Combine all the solutions

t = arctan (2 + 3) + πn, t = arctan (2 - 3) + πn

Show solutions in decimal form

t = 1.30899 \dots + πn, t = 0.26179 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2t)+tan(t)=1-tan(2t)tan(t) ?
The general solution for tan(2t)+tan(t)=1-tan(2t)tan(t) is t=1.30899…+pin,t=0.26179…+pin