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

The general solution for 1+cot^2(a)=tan^2(a) is a=0.90455…+pin,a=2.23703…+pin

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

Solution

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

Solution

a = 0.90455 \dots + πn, a = 2.23703 \dots + πn

Degrees

a = 51.82729 \dots^{\circ} + 18 0^{\circ} n, a = 128.17270 \dots^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

tan^{2} (a)

from both sides

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Solve by substitution

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

Let:

cot (a) = u

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

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

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

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

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 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 \cdot 1}

Separate the solutions

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

v = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

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

Multiply the numbers:

2 \cdot 1 = 2

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

v = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

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

Multiply the numbers:

2 \cdot 1 = 2

= \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 = \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}

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

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

Substitute back

u = cot (a)

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

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

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

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

Apply trig inverse properties

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

General solutions for

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

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

a = arccot \frac{- 1 + 5}{2} + πn

a = arccot \frac{- 1 + 5}{2} + πn

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

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

Apply trig inverse properties

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

General solutions for

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

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

a = arccot - \frac{- 1 + 5}{2} + πn

a = arccot - \frac{- 1 + 5}{2} + πn

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

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

Apply trig inverse properties

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

General solutions for

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

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

a = arccot \frac{- 1 - 5}{2} + πn

a = arccot \frac{- 1 - 5}{2} + πn

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

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

Apply trig inverse properties

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

General solutions for

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

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

a = arccot - \frac{- 1 - 5}{2} + πn

a = arccot - \frac{- 1 - 5}{2} + πn

Combine all the solutions

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

Since the equation is undefined for:

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

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

Show solutions in decimal form

a = 0.90455 \dots + πn, a = 2.23703 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 1+cot^2(a)=tan^2(a) ?
The general solution for 1+cot^2(a)=tan^2(a) is a=0.90455…+pin,a=2.23703…+pin