What is the general solution for tan(b+15)*cot(2b-5)=1 ?

The general solution for tan(b+15)*cot(2b-5)=1 is b=-2pin+20,b=-pi+20-2pin

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tan(b+15)*cot(2b-5)=1

Solution

tan (b + 15) \cdot cot (2 b - 5) = 1

Solution

b = - 2 πn + 20, b = - π + 20 - 2 πn

Degrees

b = 1145.91559 \dots^{\circ} - 36 0^{\circ} n, b = 965.91559 \dots^{\circ} - 36 0^{\circ} n

Solution steps

tan (b + 15) cot (2 b - 5) = 1

Subtract

1

from both sides

tan (b + 15) cot (2 b - 5) - 1 = 0

Express with sin, cos

- 1 + cot (- 5 + 2 b) tan (15 + b)

Use the basic trigonometric identity:

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

= - 1 + \frac{cos ( - 5 + 2 b )}{sin ( - 5 + 2 b )} tan (15 + b)

Use the basic trigonometric identity:

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

= - 1 + \frac{cos ( - 5 + 2 b )}{sin ( - 5 + 2 b )} \cdot \frac{sin ( 15 + b )}{cos ( 15 + b )}

Simplify

- 1 + \frac{cos ( - 5 + 2 b )}{sin ( - 5 + 2 b )} \cdot \frac{sin ( 15 + b )}{cos ( 15 + b )} : \frac{- sin ( - 5 + 2 b ) cos ( 15 + b ) + cos ( - 5 + 2 b ) sin ( 15 + b )}{sin ( - 5 + 2 b ) cos ( 15 + b )}

- 1 + \frac{cos ( - 5 + 2 b )}{sin ( - 5 + 2 b )} \cdot \frac{sin ( 15 + b )}{cos ( 15 + b )}

Multiply

\frac{cos ( - 5 + 2 b )}{sin ( - 5 + 2 b )} \cdot \frac{sin ( 15 + b )}{cos ( 15 + b )} : \frac{cos ( 2 b - 5 ) sin ( b + 15 )}{sin ( 2 b - 5 ) cos ( b + 15 )}

\frac{cos ( - 5 + 2 b )}{sin ( - 5 + 2 b )} \cdot \frac{sin ( 15 + b )}{cos ( 15 + b )}

Multiply fractions:

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

= \frac{cos ( - 5 + 2 b ) sin ( 15 + b )}{sin ( - 5 + 2 b ) cos ( 15 + b )}

= - 1 + \frac{cos ( 2 b - 5 ) sin ( b + 15 )}{sin ( 2 b - 5 ) cos ( b + 15 )}

Convert element to fraction:

1 = \frac{1 sin ( - 5 + 2 b ) cos ( 15 + b )}{sin ( - 5 + 2 b ) cos ( 15 + b )}

= - \frac{1 \cdot sin ( - 5 + 2 b ) cos ( 15 + b )}{sin ( - 5 + 2 b ) cos ( 15 + b )} + \frac{cos ( - 5 + 2 b ) sin ( 15 + b )}{sin ( - 5 + 2 b ) cos ( 15 + b )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot sin ( - 5 + 2 b ) cos ( 15 + b ) + cos ( - 5 + 2 b ) sin ( 15 + b )}{sin ( - 5 + 2 b ) cos ( 15 + b )}

Multiply:

1 \cdot sin (- 5 + 2 b) = sin (- 5 + 2 b)

= \frac{- sin ( 2 b - 5 ) cos ( b + 15 ) + cos ( 2 b - 5 ) sin ( b + 15 )}{sin ( 2 b - 5 ) cos ( b + 15 )}

= \frac{- sin ( - 5 + 2 b ) cos ( 15 + b ) + cos ( - 5 + 2 b ) sin ( 15 + b )}{sin ( - 5 + 2 b ) cos ( 15 + b )}

\frac{cos ( - 5 + 2 b ) sin ( 15 + b ) - cos ( 15 + b ) sin ( - 5 + 2 b )}{cos ( 15 + b ) sin ( - 5 + 2 b )} = 0

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

cos (- 5 + 2 b) sin (15 + b) - cos (15 + b) sin (- 5 + 2 b) = 0

Rewrite using trig identities

cos (- 5 + 2 b) sin (15 + b) - cos (15 + b) sin (- 5 + 2 b)

Use the Angle Difference identity:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= sin (15 + b - (- 5 + 2 b))

sin (15 + b - (- 5 + 2 b)) = 0

General solutions for

sin (15 + b - (- 5 + 2 b)) = 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}

15 + b - (- 5 + 2 b) = 0 + 2 πn, 15 + b - (- 5 + 2 b) = π + 2 πn

15 + b - (- 5 + 2 b) = 0 + 2 πn, 15 + b - (- 5 + 2 b) = π + 2 πn

Solve

15 + b - (- 5 + 2 b) = 0 + 2 πn : b = - 2 πn + 20

15 + b - (- 5 + 2 b) = 0 + 2 πn

0 + 2 πn = 2 πn

15 + b - (- 5 + 2 b) = 2 πn

Move

15

to the right side

15 + b - (- 5 + 2 b) = 2 πn

Subtract

15

from both sides

15 + b - (- 5 + 2 b) - 15 = 2 πn - 15

Simplify

b - (- 5 + 2 b) = 2 πn - 15

b - (- 5 + 2 b) = 2 πn - 15

Expand

b - (- 5 + 2 b) : - b + 5

b - (- 5 + 2 b)

- (- 5 + 2 b) : 5 - 2 b

- (- 5 + 2 b)

Distribute parentheses

= - (- 5) - (2 b)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= 5 - 2 b

= b + 5 - 2 b

Simplify

b + 5 - 2 b : - b + 5

b + 5 - 2 b

Group like terms

= b - 2 b + 5

Add similar elements:

b - 2 b = - b

= - b + 5

= - b + 5

- b + 5 = 2 πn - 15

Move

5

to the right side

- b + 5 = 2 πn - 15

Subtract

5

from both sides

- b + 5 - 5 = 2 πn - 15 - 5

Simplify

- b = 2 πn - 20

- b = 2 πn - 20

Divide both sides by

- 1

- b = 2 πn - 20

Divide both sides by

- 1

\frac{- b}{- 1} = \frac{2 πn}{- 1} - \frac{20}{- 1}

Simplify

\frac{- b}{- 1} = \frac{2 πn}{- 1} - \frac{20}{- 1}

Simplify

\frac{- b}{- 1} : b

\frac{- b}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{b}{1}

Apply rule

\frac{a}{1} = a

= b

Simplify

\frac{2 πn}{- 1} - \frac{20}{- 1} : - 2 πn + 20

\frac{2 πn}{- 1} - \frac{20}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - 2 πn - \frac{20}{- 1}

\frac{20}{- 1} = - 20

\frac{20}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{20}{1}

Apply rule

\frac{a}{1} = a

= - 20

= - 2 πn - (- 20)

Apply rule

- (- a) = a

= - 2 πn + 20

b = - 2 πn + 20

b = - 2 πn + 20

b = - 2 πn + 20

Solve

15 + b - (- 5 + 2 b) = π + 2 πn : b = - π + 20 - 2 πn

15 + b - (- 5 + 2 b) = π + 2 πn

Move

15

to the right side

15 + b - (- 5 + 2 b) = π + 2 πn

Subtract

15

from both sides

15 + b - (- 5 + 2 b) - 15 = π + 2 πn - 15

Simplify

b - (- 5 + 2 b) = π + 2 πn - 15

b - (- 5 + 2 b) = π + 2 πn - 15

Expand

b - (- 5 + 2 b) : - b + 5

b - (- 5 + 2 b)

- (- 5 + 2 b) : 5 - 2 b

- (- 5 + 2 b)

Distribute parentheses

= - (- 5) - (2 b)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= 5 - 2 b

= b + 5 - 2 b

Simplify

b + 5 - 2 b : - b + 5

b + 5 - 2 b

Group like terms

= b - 2 b + 5

Add similar elements:

b - 2 b = - b

= - b + 5

= - b + 5

- b + 5 = π + 2 πn - 15

Move

5

to the right side

- b + 5 = π + 2 πn - 15

Subtract

5

from both sides

- b + 5 - 5 = π + 2 πn - 15 - 5

Simplify

- b = π + 2 πn - 20

- b = π + 2 πn - 20

Divide both sides by

- 1

- b = π + 2 πn - 20

Divide both sides by

- 1

\frac{- b}{- 1} = \frac{π}{- 1} + \frac{2 πn}{- 1} - \frac{20}{- 1}

Simplify

\frac{- b}{- 1} = \frac{π}{- 1} + \frac{2 πn}{- 1} - \frac{20}{- 1}

Simplify

\frac{- b}{- 1} : b

\frac{- b}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{b}{1}

Apply rule

\frac{a}{1} = a

= b

Simplify

\frac{π}{- 1} + \frac{2 πn}{- 1} - \frac{20}{- 1} : - π + 20 - 2 πn

\frac{π}{- 1} + \frac{2 πn}{- 1} - \frac{20}{- 1}

Group like terms

= \frac{π}{- 1} - \frac{20}{- 1} + \frac{2 πn}{- 1}

\frac{π}{- 1} = - π

\frac{π}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{π}{1}

Apply rule

\frac{a}{1} = a

= - π

= - π - \frac{20}{- 1} + \frac{2 πn}{- 1}

\frac{20}{- 1} = - 20

\frac{20}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{20}{1}

Apply rule

\frac{a}{1} = a

= - 20

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - π - (- 20) - 2 πn

Apply rule

- (- a) = a

= - π + 20 - 2 πn

b = - π + 20 - 2 πn

b = - π + 20 - 2 πn

b = - π + 20 - 2 πn

b = - 2 πn + 20, b = - π + 20 - 2 πn

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sin (x) = - cos (2 x)

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Frequently Asked Questions (FAQ)

What is the general solution for tan(b+15)*cot(2b-5)=1 ?
The general solution for tan(b+15)*cot(2b-5)=1 is b=-2pin+20,b=-pi+20-2pin