What is the general solution for tan(8b)=cot(10b) ?

The general solution for tan(8b)=cot(10b) is b= pi/(36)+(pin)/9 ,b= pi/(12)+(pin)/9

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tan(8b)=cot(10b)

Solution

tan (8 b) = cot (10 b)

Solution

b = \frac{π}{36} + \frac{πn}{9}, b = \frac{π}{12} + \frac{πn}{9}

Degrees

b = 5^{\circ} + 2 0^{\circ} n, b = 1 5^{\circ} + 2 0^{\circ} n

Solution steps

tan (8 b) = cot (10 b)

Subtract

cot (10 b)

from both sides

tan (8 b) - cot (10 b) = 0

Express with sin, cos

- cot (10 b) + tan (8 b)

Use the basic trigonometric identity:

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

= - \frac{cos ( 10 b )}{sin ( 10 b )} + tan (8 b)

Use the basic trigonometric identity:

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

= - \frac{cos ( 10 b )}{sin ( 10 b )} + \frac{sin ( 8 b )}{cos ( 8 b )}

Simplify

- \frac{cos ( 10 b )}{sin ( 10 b )} + \frac{sin ( 8 b )}{cos ( 8 b )} : \frac{- cos ( 10 b ) cos ( 8 b ) + sin ( 8 b ) sin ( 10 b )}{sin ( 10 b ) cos ( 8 b )}

- \frac{cos ( 10 b )}{sin ( 10 b )} + \frac{sin ( 8 b )}{cos ( 8 b )}

Least Common Multiplier of

sin (10 b), cos (8 b) : sin (10 b) cos (8 b)

sin (10 b), cos (8 b)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (10 b)

cos (8 b)

= sin (10 b) cos (8 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 (10 b) cos (8 b)

For

\frac{cos ( 10 b )}{sin ( 10 b )} :

multiply the denominator and numerator by

cos (8 b)

\frac{cos ( 10 b )}{sin ( 10 b )} = \frac{cos ( 10 b ) cos ( 8 b )}{sin ( 10 b ) cos ( 8 b )}

For

\frac{sin ( 8 b )}{cos ( 8 b )} :

multiply the denominator and numerator by

sin (10 b)

\frac{sin ( 8 b )}{cos ( 8 b )} = \frac{sin ( 8 b ) sin ( 10 b )}{cos ( 8 b ) sin ( 10 b )}

= - \frac{cos ( 10 b ) cos ( 8 b )}{sin ( 10 b ) cos ( 8 b )} + \frac{sin ( 8 b ) sin ( 10 b )}{cos ( 8 b ) sin ( 10 b )}

Since the denominators are equal, combine the fractions:

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

= \frac{- cos ( 10 b ) cos ( 8 b ) + sin ( 8 b ) sin ( 10 b )}{sin ( 10 b ) cos ( 8 b )}

= \frac{- cos ( 10 b ) cos ( 8 b ) + sin ( 8 b ) sin ( 10 b )}{sin ( 10 b ) cos ( 8 b )}

\frac{- cos ( 10 b ) cos ( 8 b ) + sin ( 10 b ) sin ( 8 b )}{cos ( 8 b ) sin ( 10 b )} = 0

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

- cos (10 b) cos (8 b) + sin (10 b) sin (8 b) = 0

Rewrite using trig identities

- cos (10 b) cos (8 b) + sin (10 b) sin (8 b)

Use the Angle Sum identity:

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

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

= - cos (10 b + 8 b)

- cos (10 b + 8 b) = 0

Divide both sides by

- 1

- cos (10 b + 8 b) = 0

Divide both sides by

- 1

\frac{- cos ( 10 b + 8 b )}{- 1} = \frac{0}{- 1}

Simplify

cos (10 b + 8 b) = 0

cos (10 b + 8 b) = 0

General solutions for

cos (10 b + 8 b) = 0

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}

10 b + 8 b = \frac{π}{2} + 2 πn, 10 b + 8 b = \frac{3 π}{2} + 2 πn

10 b + 8 b = \frac{π}{2} + 2 πn, 10 b + 8 b = \frac{3 π}{2} + 2 πn

Solve

10 b + 8 b = \frac{π}{2} + 2 πn : b = \frac{π}{36} + \frac{πn}{9}

10 b + 8 b = \frac{π}{2} + 2 πn

Add similar elements:

10 b + 8 b = 18 b

18 b = \frac{π}{2} + 2 πn

Divide both sides by

18

18 b = \frac{π}{2} + 2 πn

Divide both sides by

18

\frac{18 b}{18} = \frac{\frac{π}{2}}{18} + \frac{2 πn}{18}

Simplify

\frac{18 b}{18} = \frac{\frac{π}{2}}{18} + \frac{2 πn}{18}

Simplify

\frac{18 b}{18} : b

\frac{18 b}{18}

Divide the numbers:

\frac{18}{18} = 1

= b

Simplify

\frac{\frac{π}{2}}{18} + \frac{2 πn}{18} : \frac{π}{36} + \frac{πn}{9}

\frac{\frac{π}{2}}{18} + \frac{2 πn}{18}

\frac{\frac{π}{2}}{18} = \frac{π}{36}

\frac{\frac{π}{2}}{18}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 18}

Multiply the numbers:

2 \cdot 18 = 36

= \frac{π}{36}

\frac{2 πn}{18} = \frac{πn}{9}

\frac{2 πn}{18}

Cancel the common factor:

2

= \frac{πn}{9}

= \frac{π}{36} + \frac{πn}{9}

b = \frac{π}{36} + \frac{πn}{9}

b = \frac{π}{36} + \frac{πn}{9}

b = \frac{π}{36} + \frac{πn}{9}

Solve

10 b + 8 b = \frac{3 π}{2} + 2 πn : b = \frac{π}{12} + \frac{πn}{9}

10 b + 8 b = \frac{3 π}{2} + 2 πn

Add similar elements:

10 b + 8 b = 18 b

18 b = \frac{3 π}{2} + 2 πn

Divide both sides by

18

18 b = \frac{3 π}{2} + 2 πn

Divide both sides by

18

\frac{18 b}{18} = \frac{\frac{3 π}{2}}{18} + \frac{2 πn}{18}

Simplify

\frac{18 b}{18} = \frac{\frac{3 π}{2}}{18} + \frac{2 πn}{18}

Simplify

\frac{18 b}{18} : b

\frac{18 b}{18}

Divide the numbers:

\frac{18}{18} = 1

= b

Simplify

\frac{\frac{3 π}{2}}{18} + \frac{2 πn}{18} : \frac{π}{12} + \frac{πn}{9}

\frac{\frac{3 π}{2}}{18} + \frac{2 πn}{18}

\frac{\frac{3 π}{2}}{18} = \frac{π}{12}

\frac{\frac{3 π}{2}}{18}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 18}

Multiply the numbers:

2 \cdot 18 = 36

= \frac{3 π}{36}

Cancel the common factor:

3

= \frac{π}{12}

\frac{2 πn}{18} = \frac{πn}{9}

\frac{2 πn}{18}

Cancel the common factor:

2

= \frac{πn}{9}

= \frac{π}{12} + \frac{πn}{9}

b = \frac{π}{12} + \frac{πn}{9}

b = \frac{π}{12} + \frac{πn}{9}

b = \frac{π}{12} + \frac{πn}{9}

b = \frac{π}{36} + \frac{πn}{9}, b = \frac{π}{12} + \frac{πn}{9}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(8b)=cot(10b) ?
The general solution for tan(8b)=cot(10b) is b= pi/(36)+(pin)/9 ,b= pi/(12)+(pin)/9