What is the general solution for (1+csc(γ))/(cot(γ)+cos(γ))=csc(γ) ?

The general solution for (1+csc(γ))/(cot(γ)+cos(γ))=csc(γ) is γ= pi/4+pin

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(1+csc(γ))/(cot(γ)+cos(γ))=csc(γ)

Solution

\frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} = csc (γ)

Solution

γ = \frac{π}{4} + πn

Degrees

γ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

\frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} = csc (γ)

Subtract

csc (γ)

from both sides

\frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} - csc (γ) = 0

Simplify

\frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} - csc (γ) : \frac{1 + csc ( γ ) - csc ( γ ) ( cot ( γ ) + cos ( γ ) )}{cot ( γ ) + cos ( γ )}

\frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} - csc (γ)

Convert element to fraction:

csc (γ) = \frac{csc ( γ ) ( cot ( γ ) + cos ( γ ) )}{cot ( γ ) + cos ( γ )}

= \frac{1 + csc ( γ )}{cot ( γ ) + cos ( γ )} - \frac{csc ( γ ) ( cot ( γ ) + cos ( γ ) )}{cot ( γ ) + cos ( γ )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + csc ( γ ) - csc ( γ ) ( cot ( γ ) + cos ( γ ) )}{cot ( γ ) + cos ( γ )}

\frac{1 + csc ( γ ) - csc ( γ ) ( cot ( γ ) + cos ( γ ) )}{cot ( γ ) + cos ( γ )} = 0

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

1 + csc (γ) - csc (γ) (cot (γ) + cos (γ)) = 0

Express with sin, cos

1 + csc (γ) - (cos (γ) + cot (γ)) csc (γ)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= 1 + \frac{1}{sin ( γ )} - (cos (γ) + cot (γ)) \frac{1}{sin ( γ )}

Use the basic trigonometric identity:

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

= 1 + \frac{1}{sin ( γ )} - (cos (γ) + \frac{cos ( γ )}{sin ( γ )}) \frac{1}{sin ( γ )}

Simplify

1 + \frac{1}{sin ( γ )} - (cos (γ) + \frac{cos ( γ )}{sin ( γ )}) \frac{1}{sin ( γ )} : \frac{sin ^{2} ( γ ) + sin ( γ ) - cos ( γ ) sin ( γ ) - cos ( γ )}{sin ^{2} ( γ )}

1 + \frac{1}{sin ( γ )} - (cos (γ) + \frac{cos ( γ )}{sin ( γ )}) \frac{1}{sin ( γ )}

(cos (γ) + \frac{cos ( γ )}{sin ( γ )}) \frac{1}{sin ( γ )} = \frac{cos ( γ ) sin ( γ ) + cos ( γ )}{sin ^{2} ( γ )}

(cos (γ) + \frac{cos ( γ )}{sin ( γ )}) \frac{1}{sin ( γ )}

Multiply fractions:

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

= \frac{1 \cdot ( cos ( γ ) + \frac{c o s ( γ )}{s i n ( γ )} )}{sin ( γ )}

1 \cdot (cos (γ) + \frac{cos ( γ )}{sin ( γ )}) = cos (γ) + \frac{cos ( γ )}{sin ( γ )}

1 \cdot (cos (γ) + \frac{cos ( γ )}{sin ( γ )})

Multiply:

1 \cdot (cos (γ) + \frac{cos ( γ )}{sin ( γ )}) = (cos (γ) + \frac{cos ( γ )}{sin ( γ )})

= (cos (γ) + \frac{cos ( γ )}{sin ( γ )})

Remove parentheses:

(a) = a

= cos (γ) + \frac{cos ( γ )}{sin ( γ )}

= \frac{cos ( γ ) + \frac{c o s ( γ )}{s i n ( γ )}}{sin ( γ )}

Join

cos (γ) + \frac{cos ( γ )}{sin ( γ )} : \frac{cos ( γ ) sin ( γ ) + cos ( γ )}{sin ( γ )}

cos (γ) + \frac{cos ( γ )}{sin ( γ )}

Convert element to fraction:

cos (γ) = \frac{cos ( γ ) sin ( γ )}{sin ( γ )}

= \frac{cos ( γ ) sin ( γ )}{sin ( γ )} + \frac{cos ( γ )}{sin ( γ )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ( γ ) sin ( γ ) + cos ( γ )}{sin ( γ )}

= \frac{\frac{c o s ( γ ) s i n ( γ ) + c o s ( γ )}{s i n ( γ )}}{sin ( γ )}

Apply the fraction rule:

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

= \frac{cos ( γ ) sin ( γ ) + cos ( γ )}{sin ( γ ) sin ( γ )}

sin (γ) sin (γ) = sin^{2} (γ)

sin (γ) sin (γ)

Apply exponent rule:

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

sin (γ) sin (γ) = sin^{1 + 1} (γ)

= sin^{1 + 1} (γ)

Add the numbers:

1 + 1 = 2

= sin^{2} (γ)

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

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

Convert element to fraction:

1 = \frac{1}{1}

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

Least Common Multiplier of

1, sin (γ), sin^{2} (γ) : sin^{2} (γ)

1, sin (γ), sin^{2} (γ)

Lowest Common Multiplier (LCM)

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

= sin^{2} (γ)

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

sin^{2} (γ)

For

\frac{1}{1} :

multiply the denominator and numerator by

sin^{2} (γ)

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

For

\frac{1}{sin ( γ )} :

multiply the denominator and numerator by

sin (γ)

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

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

Since the denominators are equal, combine the fractions:

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

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

- (cos (γ) sin (γ) + cos (γ)) : - cos (γ) sin (γ) - cos (γ)

- (cos (γ) sin (γ) + cos (γ))

Distribute parentheses

= - (cos (γ) sin (γ)) - (cos (γ))

Apply minus-plus rules

+ (- a) = - a

= - cos (γ) sin (γ) - cos (γ)

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

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

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

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

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

Factor

- cos (γ) + sin (γ) + sin^{2} (γ) - cos (γ) sin (γ) : (1 + sin (γ)) (- cos (γ) + sin (γ))

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

Factor out common term

cos (γ)

= - cos (γ) (1 + sin (γ)) + sin (γ) + sin^{2} (γ)

Apply exponent rule:

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

sin^{2} (γ) = sin (γ) sin (γ)

= - cos (γ) (1 + sin (γ)) + sin (γ) + sin (γ) sin (γ)

Factor out common term

sin (γ)

= - cos (γ) (1 + sin (γ)) + sin (γ) (1 + sin (γ))

Factor out common term

(1 + sin (γ))

= (1 + sin (γ)) (- cos (γ) + sin (γ))

(1 + sin (γ)) (- cos (γ) + sin (γ)) = 0

Solving each part separately

1 + sin (γ) = 0 or - cos (γ) + sin (γ) = 0

1 + sin (γ) = 0 : γ = \frac{3 π}{2} + 2 πn

1 + sin (γ) = 0

Move

1

to the right side

1 + sin (γ) = 0

Subtract

1

from both sides

1 + sin (γ) - 1 = 0 - 1

Simplify

sin (γ) = - 1

sin (γ) = - 1

General solutions for

sin (γ) = - 1

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}

γ = \frac{3 π}{2} + 2 πn

γ = \frac{3 π}{2} + 2 πn

- cos (γ) + sin (γ) = 0 : γ = \frac{π}{4} + πn

- cos (γ) + sin (γ) = 0

Rewrite using trig identities

- cos (γ) + sin (γ) = 0

Divide both sides by

cos (γ), cos (γ) \neq = 0

\frac{- cos ( γ ) + sin ( γ )}{cos ( γ )} = \frac{0}{cos ( γ )}

Simplify

- 1 + \frac{sin ( γ )}{cos ( γ )} = 0

Use the basic trigonometric identity:

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

- 1 + tan (γ) = 0

- 1 + tan (γ) = 0

Move

1

to the right side

- 1 + tan (γ) = 0

Add

1

to both sides

- 1 + tan (γ) + 1 = 0 + 1

Simplify

tan (γ) = 1

tan (γ) = 1

General solutions for

tan (γ) = 1

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}

γ = \frac{π}{4} + πn

γ = \frac{π}{4} + πn

Combine all the solutions

γ = \frac{3 π}{2} + 2 πn, γ = \frac{π}{4} + πn

Since the equation is undefined for:

\frac{3 π}{2} + 2 πn

γ = \frac{π}{4} + πn

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tan(x)=(7.3)/(6.8)

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Frequently Asked Questions (FAQ)

What is the general solution for (1+csc(γ))/(cot(γ)+cos(γ))=csc(γ) ?
The general solution for (1+csc(γ))/(cot(γ)+cos(γ))=csc(γ) is γ= pi/4+pin