What is the general solution for 1/(tan(α))+tan(α)= 1/(sin(α)) ?

The general solution for 1/(tan(α))+tan(α)= 1/(sin(α)) is No Solution for α\in\mathbb{R}

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1/(tan(α))+tan(α)= 1/(sin(α))

Solution

\frac{1}{tan ( α )} + tan (α) = \frac{1}{sin ( α )}

Solution

No Solution for α \in R

Solution steps

\frac{1}{tan ( α )} + tan (α) = \frac{1}{sin ( α )}

Subtract

\frac{1}{sin ( α )}

from both sides

\frac{1}{tan ( α )} + tan (α) - \frac{1}{sin ( α )} = 0

Simplify

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

\frac{1}{tan ( α )} + tan (α) - \frac{1}{sin ( α )}

Convert element to fraction:

tan (α) = \frac{tan ( α )}{1}

= \frac{1}{tan ( α )} + \frac{tan ( α )}{1} - \frac{1}{sin ( α )}

Least Common Multiplier of

tan (α), 1, sin (α) : tan (α) sin (α)

tan (α), 1, sin (α)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear in at least one of the factored expressions

= tan (α) sin (α)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

tan (α) sin (α)

For

\frac{1}{tan ( α )} :

multiply the denominator and numerator by

sin (α)

\frac{1}{tan ( α )} = \frac{1 \cdot sin ( α )}{tan ( α ) sin ( α )} = \frac{sin ( α )}{tan ( α ) sin ( α )}

For

\frac{tan ( α )}{1} :

multiply the denominator and numerator by

tan (α) sin (α)

\frac{tan ( α )}{1} = \frac{tan ( α ) tan ( α ) sin ( α )}{1 \cdot tan ( α ) sin ( α )} = \frac{tan ^{2} ( α ) sin ( α )}{tan ( α ) sin ( α )}

For

\frac{1}{sin ( α )} :

multiply the denominator and numerator by

tan (α)

\frac{1}{sin ( α )} = \frac{1 \cdot tan ( α )}{sin ( α ) tan ( α )} = \frac{tan ( α )}{tan ( α ) sin ( α )}

= \frac{sin ( α )}{tan ( α ) sin ( α )} + \frac{tan ^{2} ( α ) sin ( α )}{tan ( α ) sin ( α )} - \frac{tan ( α )}{tan ( α ) sin ( α )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( α ) + tan ^{2} ( α ) sin ( α ) - tan ( α )}{tan ( α ) sin ( α )}

\frac{sin ( α ) + tan ^{2} ( α ) sin ( α ) - tan ( α )}{tan ( α ) sin ( α )} = 0

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

sin (α) + tan^{2} (α) sin (α) - tan (α) = 0

Express with sin, cos

sin (α) - tan (α) + sin (α) tan^{2} (α)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= sin (α) - \frac{sin ( α )}{cos ( α )} + sin (α) (\frac{sin ( α )}{cos ( α )})^{2}

Simplify

sin (α) - \frac{sin ( α )}{cos ( α )} + sin (α) (\frac{sin ( α )}{cos ( α )})^{2} : \frac{cos ^{2} ( α ) sin ( α ) - sin ( α ) cos ( α ) + sin ^{3} ( α )}{cos ^{2} ( α )}

sin (α) - \frac{sin ( α )}{cos ( α )} + sin (α) (\frac{sin ( α )}{cos ( α )})^{2}

sin (α) (\frac{sin ( α )}{cos ( α )})^{2} = \frac{sin ^{3} ( α )}{cos ^{2} ( α )}

sin (α) (\frac{sin ( α )}{cos ( α )})^{2}

(\frac{sin ( α )}{cos ( α )})^{2} = \frac{sin ^{2} ( α )}{cos ^{2} ( α )}

(\frac{sin ( α )}{cos ( α )})^{2}

Apply exponent rule:

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

= \frac{sin ^{2} ( α )}{cos ^{2} ( α )}

= \frac{sin ^{2} ( α )}{cos ^{2} ( α )} sin (α)

Multiply fractions:

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

= \frac{sin ^{2} ( α ) sin ( α )}{cos ^{2} ( α )}

sin^{2} (α) sin (α) = sin^{3} (α)

sin^{2} (α) sin (α)

Apply exponent rule:

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

sin^{2} (α) sin (α) = sin^{2 + 1} (α)

= sin^{2 + 1} (α)

Add the numbers:

2 + 1 = 3

= sin^{3} (α)

= \frac{sin ^{3} ( α )}{cos ^{2} ( α )}

= sin (α) - \frac{sin ( α )}{cos ( α )} + \frac{sin ^{3} ( α )}{cos ^{2} ( α )}

Convert element to fraction:

sin (α) = \frac{sin ( α )}{1}

= \frac{sin ( α )}{1} - \frac{sin ( α )}{cos ( α )} + \frac{sin ^{3} ( α )}{cos ^{2} ( α )}

Least Common Multiplier of

1, cos (α), cos^{2} (α) : cos^{2} (α)

1, cos (α), cos^{2} (α)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear in at least one of the factored expressions

= cos^{2} (α)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos^{2} (α)

For

\frac{sin ( α )}{1} :

multiply the denominator and numerator by

cos^{2} (α)

\frac{sin ( α )}{1} = \frac{sin ( α ) cos ^{2} ( α )}{1 \cdot cos ^{2} ( α )} = \frac{sin ( α ) cos ^{2} ( α )}{cos ^{2} ( α )}

For

\frac{sin ( α )}{cos ( α )} :

multiply the denominator and numerator by

cos (α)

\frac{sin ( α )}{cos ( α )} = \frac{sin ( α ) cos ( α )}{cos ( α ) cos ( α )} = \frac{sin ( α ) cos ( α )}{cos ^{2} ( α )}

= \frac{sin ( α ) cos ^{2} ( α )}{cos ^{2} ( α )} - \frac{sin ( α ) cos ( α )}{cos ^{2} ( α )} + \frac{sin ^{3} ( α )}{cos ^{2} ( α )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( α ) cos ^{2} ( α ) - sin ( α ) cos ( α ) + sin ^{3} ( α )}{cos ^{2} ( α )}

= \frac{cos ^{2} ( α ) sin ( α ) - sin ( α ) cos ( α ) + sin ^{3} ( α )}{cos ^{2} ( α )}

\frac{sin ^{3} ( α ) - cos ( α ) sin ( α ) + cos ^{2} ( α ) sin ( α )}{cos ^{2} ( α )} = 0

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

sin^{3} (α) - cos (α) sin (α) + cos^{2} (α) sin (α) = 0

Factor

sin^{3} (α) - cos (α) sin (α) + cos^{2} (α) sin (α) : sin (α) (sin^{2} (α) - cos (α) + cos^{2} (α))

sin^{3} (α) - cos (α) sin (α) + cos^{2} (α) sin (α)

Apply exponent rule:

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

sin^{3} (α) = sin (α) sin^{2} (α)

= sin (α) sin^{2} (α) - sin (α) cos (α) + sin (α) cos^{2} (α)

Factor out common term

sin (α)

= sin (α) (sin^{2} (α) - cos (α) + cos^{2} (α))

sin (α) (sin^{2} (α) - cos (α) + cos^{2} (α)) = 0

Rewrite using trig identities

sin (α) (sin^{2} (α) - cos (α) + cos^{2} (α))

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= sin (α) (- cos (α) + 1)

sin (α) (- cos (α) + 1) = 0

Solving each part separately

sin (α) = 0 or - cos (α) + 1 = 0

sin (α) = 0 : α = 2 πn, α = π + 2 πn

sin (α) = 0

General solutions for

sin (α) = 0

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

α = 0 + 2 πn, α = π + 2 πn

α = 0 + 2 πn, α = π + 2 πn

Solve

α = 0 + 2 πn : α = 2 πn

α = 0 + 2 πn

0 + 2 πn = 2 πn

α = 2 πn

α = 2 πn, α = π + 2 πn

- cos (α) + 1 = 0 : α = 2 πn

- cos (α) + 1 = 0

Move

1

to the right side

- cos (α) + 1 = 0

Subtract

1

from both sides

- cos (α) + 1 - 1 = 0 - 1

Simplify

- cos (α) = - 1

- cos (α) = - 1

Divide both sides by

- 1

- cos (α) = - 1

Divide both sides by

- 1

\frac{- cos ( α )}{- 1} = \frac{- 1}{- 1}

Simplify

cos (α) = 1

cos (α) = 1

General solutions for

cos (α) = 1

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

α = 0 + 2 πn

α = 0 + 2 πn

Solve

α = 0 + 2 πn : α = 2 πn

α = 0 + 2 πn

0 + 2 πn = 2 πn

α = 2 πn

α = 2 πn

Combine all the solutions

α = 2 πn, α = π + 2 πn

Since the equation is undefined for:

2 πn, π + 2 πn

No Solution for α \in R

Popular Examples

tan(2θ)=(-81)/(81)

tan (2 θ) = \frac{- 81}{81}

sin(θ)cos(2θ)-cos(2θ)=0

sin (θ) cos (2 θ) - cos (2 θ) = 0

tan(θ)=-7/6

tan (θ) = - \frac{7}{6}

8sin^2(x)-1=5

8 sin^{2} (x) - 1 = 5

sin(d)=0.8

sin (d) = 0.8

Frequently Asked Questions (FAQ)

What is the general solution for 1/(tan(α))+tan(α)= 1/(sin(α)) ?
The general solution for 1/(tan(α))+tan(α)= 1/(sin(α)) is No Solution for α\in\mathbb{R}