What is the general solution for tan^2(x)+cot^2(x)-2=0 ?

The general solution for tan^2(x)+cot^2(x)-2=0 is x= pi/4+pin,x=(3pi)/4+pin

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tan^2(x)+cot^2(x)-2=0

Solution

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

Solution

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Rewrite using trig identities

- 2 + cot^{2} (x) + tan^{2} (x)

Use the basic trigonometric identity:

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

Solve by substitution

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

Let:

cot (x) = u

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

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

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

Multiply both sides by

u^{2}

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

Multiply both sides by

u^{2}

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

Simplify

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

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

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

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

Solve

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

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

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} - 2 v + 1 = 0

Solve

v^{2} - 2 v + 1 = 0 : v = 1

v^{2} - 2 v + 1 = 0

Solve with the quadratic formula

v^{2} - 2 v + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 1

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

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

v_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

v = \frac{- ( - 2 )}{2 \cdot 1}

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

v = 1

The solution to the quadratic equation is:

v = 1

v = 1

Substitute back

v = u^{2},

solve for

u

Solve

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

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

The solutions are

u = 1, u = - 1

u = 1, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- 2 + 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 = 1, u = - 1

Substitute back

u = cot (x)

cot (x) = 1, cot (x) = - 1

cot (x) = 1, cot (x) = - 1

cot (x) = 1 : x = \frac{π}{4} + πn

cot (x) = 1

General solutions for

cot (x) = 1

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

cot (x) = - 1 : x = \frac{3 π}{4} + πn

cot (x) = - 1

General solutions for

cot (x) = - 1

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan^2(x)+cot^2(x)-2=0 ?
The general solution for tan^2(x)+cot^2(x)-2=0 is x= pi/4+pin,x=(3pi)/4+pin