What is the general solution for tan(2A)=((2tan(A)))/([1+(tan(A))^2]) ?

The general solution for tan(2A)=((2tan(A)))/([1+(tan(A))^2]) is A=pin

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tan(2A)=((2tan(A)))/([1+(tan(A))^2])

Solution

tan (2 A) = \frac{( 2 tan ( A ) )}{[ 1 + ( tan ( A ) ) ^{2} ]}

Solution

A = πn

Degrees

A = 0^{\circ} + 18 0^{\circ} n

Solution steps

tan (2 A) = \frac{( 2 tan ( A ) )}{[ 1 + ( tan ( A ) ) ^{2} ]}

Subtract

\frac{2 tan ( A )}{[ 1 + ( tan ( A ) ) ^{2} ]}

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

(1 + tan^{2} (A)) tan (2 A) - 2 tan (A)

Use the Double Angle identity:

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

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

Simplify

- 2 tan (A) + \frac{2 tan ( A )}{1 - tan ^{2} ( A )} (1 + tan^{2} (A)) : \frac{4 tan ^{3} ( A )}{1 - tan ^{2} ( A )}

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

Multiply

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

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

Multiply fractions:

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

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

2 tan (A) (1 + tan^{2} (A)) - 2 tan (A) (1 - tan^{2} (A))

Expand

2 tan (A) (1 + tan^{2} (A)) : 2 tan (A) + 2 tan^{3} (A)

2 tan (A) (1 + tan^{2} (A))

Apply the distributive law:

a (b + c) = ab + a c

a = 2 tan (A), b = 1, c = tan^{2} (A)

= 2 tan (A) \cdot 1 + 2 tan (A) tan^{2} (A)

= 2 \cdot 1 \cdot tan (A) + 2 tan^{2} (A) tan (A)

Simplify

2 \cdot 1 \cdot tan (A) + 2 tan^{2} (A) tan (A) : 2 tan (A) + 2 tan^{3} (A)

2 \cdot 1 \cdot tan (A) + 2 tan^{2} (A) tan (A)

2 \cdot 1 \cdot tan (A) = 2 tan (A)

2 \cdot 1 \cdot tan (A)

Multiply the numbers:

2 \cdot 1 = 2

= 2 tan (A)

2 tan^{2} (A) tan (A) = 2 tan^{3} (A)

2 tan^{2} (A) tan (A)

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 tan^{3} (A)

= 2 tan (A) + 2 tan^{3} (A)

= 2 tan (A) + 2 tan^{3} (A)

= 2 tan (A) + 2 tan^{3} (A) - 2 tan (A) (1 - tan^{2} (A))

Expand

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

- 2 tan (A) (1 - tan^{2} (A))

Apply the distributive law:

a (b - c) = ab - a c

a = - 2 tan (A), b = 1, c = tan^{2} (A)

= - 2 tan (A) \cdot 1 - (- 2 tan (A)) tan^{2} (A)

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 \cdot tan (A) + 2 tan^{2} (A) tan (A)

Simplify

- 2 \cdot 1 \cdot tan (A) + 2 tan^{2} (A) tan (A) : - 2 tan (A) + 2 tan^{3} (A)

- 2 \cdot 1 \cdot tan (A) + 2 tan^{2} (A) tan (A)

2 \cdot 1 \cdot tan (A) = 2 tan (A)

2 \cdot 1 \cdot tan (A)

Multiply the numbers:

2 \cdot 1 = 2

= 2 tan (A)

2 tan^{2} (A) tan (A) = 2 tan^{3} (A)

2 tan^{2} (A) tan (A)

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 tan^{3} (A)

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

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

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

Simplify

2 tan (A) + 2 tan^{3} (A) - 2 tan (A) + 2 tan^{3} (A) : 4 tan^{3} (A)

2 tan (A) + 2 tan^{3} (A) - 2 tan (A) + 2 tan^{3} (A)

Add similar elements:

2 tan^{3} (A) + 2 tan^{3} (A) = 4 tan^{3} (A)

= 2 tan (A) + 4 tan^{3} (A) - 2 tan (A)

Add similar elements:

2 tan (A) - 2 tan (A) = 0

= 4 tan^{3} (A)

= 4 tan^{3} (A)

= \frac{4 tan ^{3} ( A )}{1 - tan ^{2} ( A )}

= \frac{4 tan ^{3} ( A )}{1 - tan ^{2} ( A )}

\frac{4 tan ^{3} ( A )}{1 - tan ^{2} ( A )} = 0

Solve by substitution

\frac{4 tan ^{3} ( A )}{1 - tan ^{2} ( A )} = 0

Let:

tan (A) = u

\frac{4 u ^{3}}{1 - u ^{2}} = 0

\frac{4 u ^{3}}{1 - u ^{2}} = 0 : u = 0

\frac{4 u ^{3}}{1 - u ^{2}} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 u^{3} = 0

Solve

4 u^{3} = 0 : u = 0

4 u^{3} = 0

Divide both sides by

4

4 u^{3} = 0

Divide both sides by

4

4 u^{3} = 0

Divide both sides by

4

\frac{4 u ^{3}}{4} = \frac{0}{4}

Simplify

u^{3} = 0

u^{3} = 0

Apply rule

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

u = 0

u = 0

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1

Take the denominator(s) of

\frac{4 u ^{3}}{1 - u ^{2}}

and compare to zero

Solve

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

1 - u^{2} = 0

Move

1

to the right side

1 - u^{2} = 0

Subtract

1

from both sides

1 - u^{2} - 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 radical rule:

1 = 1

= 1

- 1 = - 1

- 1

Apply radical rule:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

The following points are undefined

u = 1, u = - 1

Combine undefined points with solutions:

u = 0

Substitute back

u = tan (A)

tan (A) = 0

tan (A) = 0

tan (A) = 0 : A = πn

tan (A) = 0

General solutions for

tan (A) = 0

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

A = 0 + πn

A = 0 + πn

Solve

A = 0 + πn : A = πn

A = 0 + πn

0 + πn = πn

A = πn

A = πn

Combine all the solutions

A = πn

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solvefor x,-44sin(2x)+42cos(2x)=0 solve for

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2A)=((2tan(A)))/([1+(tan(A))^2]) ?
The general solution for tan(2A)=((2tan(A)))/([1+(tan(A))^2]) is A=pin