What is the general solution for cot(θ)+tan(θ)=2 ?

The general solution for cot(θ)+tan(θ)=2 is θ= pi/4+pin

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cot(θ)+tan(θ)=2

Solution

cot (θ) + tan (θ) = 2

Solution

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

cot (θ) + tan (θ) = 2

Subtract

2

from both sides

cot (θ) + tan (θ) - 2 = 0

Rewrite using trig identities

- 2 + cot (θ) + tan (θ)

Use the basic trigonometric identity:

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

= - 2 + cot (θ) + \frac{1}{cot ( θ )}

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

Solve by substitution

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

Let:

cot (θ) = u

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

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

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

Simplify

uu : u^{2}

uu

Apply exponent rule:

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

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot u}{u}

Cancel the common factor:

u

= 1

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

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

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

u_{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

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

u = \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

u = 1

The solution to the quadratic equation is:

u = 1

u = 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1

Substitute back

u = cot (θ)

cot (θ) = 1

cot (θ) = 1

cot (θ) = 1 : θ = \frac{π}{4} + πn

cot (θ) = 1

General solutions for

cot (θ) = 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

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for cot(θ)+tan(θ)=2 ?
The general solution for cot(θ)+tan(θ)=2 is θ= pi/4+pin