What is the general solution for (sec(B)+csc(B))/(1+tan(B))=csc^2(B) ?

The general solution for (sec(B)+csc(B))/(1+tan(B))=csc^2(B) is No Solution for B\in\mathbb{R}

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Popular Trigonometry >

(sec(B)+csc(B))/(1+tan(B))=csc^2(B)

Solution

\frac{sec ( B ) + csc ( B )}{1 + tan ( B )} = csc^{2} (B)

Solution

No Solution for B \in R

Solution steps

\frac{sec ( B ) + csc ( B )}{1 + tan ( B )} = csc^{2} (B)

Subtract

csc^{2} (B)

from both sides

\frac{sec ( B ) + csc ( B )}{1 + tan ( B )} - csc^{2} (B) = 0

Simplify

\frac{sec ( B ) + csc ( B )}{1 + tan ( B )} - csc^{2} (B) : \frac{sec ( B ) + csc ( B ) - csc ^{2} ( B ) ( 1 + tan ( B ) )}{1 + tan ( B )}

\frac{sec ( B ) + csc ( B )}{1 + tan ( B )} - csc^{2} (B)

Convert element to fraction:

csc^{2} (B) = \frac{csc ^{2} ( B ) ( 1 + tan ( B ) )}{1 + tan ( B )}

= \frac{sec ( B ) + csc ( B )}{1 + tan ( B )} - \frac{csc ^{2} ( B ) ( 1 + tan ( B ) )}{1 + tan ( B )}

Since the denominators are equal, combine the fractions:

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

= \frac{sec ( B ) + csc ( B ) - csc ^{2} ( B ) ( 1 + tan ( B ) )}{1 + tan ( B )}

\frac{sec ( B ) + csc ( B ) - csc ^{2} ( B ) ( 1 + tan ( B ) )}{1 + tan ( B )} = 0

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

sec (B) + csc (B) - csc^{2} (B) (1 + tan (B)) = 0

Express with sin, cos

csc (B) + sec (B) - (1 + tan (B)) csc^{2} (B)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( B )} + sec (B) - (1 + tan (B)) (\frac{1}{sin ( B )})^{2}

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Join

1 + \frac{sin ( B )}{cos ( B )} : \frac{cos ( B ) + sin ( B )}{cos ( B )}

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

Convert element to fraction:

1 = \frac{1 cos ( B )}{cos ( B )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (B) = cos (B)

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

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

(\frac{1}{sin ( B )})^{2} = \frac{1}{sin ^{2} ( B )}

(\frac{1}{sin ( B )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( B )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( B )}

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

Multiply fractions:

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

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

(cos (B) + sin (B)) \cdot 1 = cos (B) + sin (B)

(cos (B) + sin (B)) \cdot 1

Multiply:

(cos (B) + sin (B)) \cdot 1 = (cos (B) + sin (B))

= (cos (B) + sin (B))

Remove parentheses:

(a) = a

= cos (B) + sin (B)

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

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

Least Common Multiplier of

sin (B), cos (B), cos (B) sin^{2} (B) : sin^{2} (B) cos (B)

sin (B), cos (B), cos (B) sin^{2} (B)

Lowest Common Multiplier (LCM)

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

= sin^{2} (B) cos (B)

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} (B) cos (B)

For

\frac{1}{sin ( B )} :

multiply the denominator and numerator by

sin (B) cos (B)

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

For

\frac{1}{cos ( B )} :

multiply the denominator and numerator by

sin^{2} (B)

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

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

Since the denominators are equal, combine the fractions:

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

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

- (cos (B) + sin (B)) : - cos (B) - sin (B)

- (cos (B) + sin (B))

Distribute parentheses

= - (cos (B)) - (sin (B))

Apply minus-plus rules

+ (- a) = - a

= - cos (B) - sin (B)

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

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

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

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

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

Factor

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

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

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

Factor out

sin (B)

from

sin^{2} (B) + sin (B) cos (B) : sin (B) (sin (B) + cos (B))

sin^{2} (B) + sin (B) cos (B)

Apply exponent rule:

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

sin^{2} (B) = sin (B) sin (B)

= sin (B) sin (B) + sin (B) cos (B)

Factor out common term

sin (B)

= sin (B) (sin (B) + cos (B))

Factor out

- 1

from

- sin (B) - cos (B) : - (sin (B) + cos (B))

- sin (B) - cos (B)

Factor out common term

- 1

= - (sin (B) + cos (B))

= sin (B) (sin (B) + cos (B)) - (sin (B) + cos (B))

Factor out common term

sin (B) + cos (B)

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

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

Solving each part separately

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

sin (B) + cos (B) = 0 : B = \frac{3 π}{4} + πn

sin (B) + cos (B) = 0

Rewrite using trig identities

sin (B) + cos (B) = 0

Divide both sides by

cos (B), cos (B) \neq = 0

\frac{sin ( B ) + cos ( B )}{cos ( B )} = \frac{0}{cos ( B )}

Simplify

\frac{sin ( B )}{cos ( B )} + 1 = 0

Use the basic trigonometric identity:

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

tan (B) + 1 = 0

tan (B) + 1 = 0

Move

1

to the right side

tan (B) + 1 = 0

Subtract

1

from both sides

tan (B) + 1 - 1 = 0 - 1

Simplify

tan (B) = - 1

tan (B) = - 1

General solutions for

tan (B) = - 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}

B = \frac{3 π}{4} + πn

B = \frac{3 π}{4} + πn

sin (B) - 1 = 0 : B = \frac{π}{2} + 2 πn

sin (B) - 1 = 0

Move

1

to the right side

sin (B) - 1 = 0

Add

1

to both sides

sin (B) - 1 + 1 = 0 + 1

Simplify

sin (B) = 1

sin (B) = 1

General solutions for

sin (B) = 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}

B = \frac{π}{2} + 2 πn

B = \frac{π}{2} + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

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

No Solution for B \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for (sec(B)+csc(B))/(1+tan(B))=csc^2(B) ?
The general solution for (sec(B)+csc(B))/(1+tan(B))=csc^2(B) is No Solution for B\in\mathbb{R}