Exercise 16 Page 523

To find the value of

tan (a - b),

we will start by recalling the formula for the tangent of a difference.

tan (a - b) = \frac{tan a - tan b}{1 + tan a tan b}

Therefore, we need to know the values of

tan a,

and

tan b .

To do it, recall the definition of a tangent.

tan (θ) = \frac{sin θ}{cos θ}

Therefore, to find

tan a

and

tan b,

we need to know

sin a,

cos a,

sin b

and

cos b .

We are given that

cos a = \frac{4}{5}

and

sin b = - \frac{1 5}{1 7} .

With this information we can find the values of

sin a

and

cos b .

Let's start by finding

sin a .

To do so we will use one of the Pythagorean Identities.

sin^{2} a + cos^{2} a = 1

In this identity, we can substitute

\frac{4}{5}

for

cos a

and solve for

sin a .

sin^{2} a + cos^{2} a = 1

Substitute

$cos a = \frac{4}{5}$

sin^{2} a + (\frac{4}{5})^{2} = 1

▼

Solve for

sin a

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

sin^{2} a + \frac{1 6}{2 5} = 1

SubEqn

$LHS - \frac{1 6}{2 5} = RHS - \frac{1 6}{2 5}$

sin^{2} a = 1 - \frac{1 6}{2 5}

OneToFrac

Rewrite $1$ as $\frac{2 5}{2 5}$

sin^{2} a = \frac{2 5}{2 5} - \frac{1 6}{2 5}

SubFrac

Subtract fractions

sin^{2} a = \frac{9}{2 5}

SqrtEqn

$LHS = RHS$

sin a = \pm \frac{9}{2 5}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

sin a = \pm \frac{3}{5}

Let's now determine the sign of

sin a .

We are told that

a

is greater than

0

and less than

\frac{π}{2} .

Therefore, if we draw angle

a

in standard position, its terminal side will be located in the first quadrant.

If the terminal side of an angle in standard position is in the first quadrant, then its sine is positive. Therefore, we have that

sin a = \frac{3}{5} .

With this in mind, we can evaluate

tan a .

tan a = \frac{sin a}{cos a}

SubstituteII

$sin a = \frac{3}{5}$ , $cos a = \frac{4}{5}$

tan a = \frac{3}{5} / \frac{4}{5}

▼

Solve for

tan a

DivFracByFracSL

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

tan a = \frac{3}{5} \cdot \frac{5}{4}

MultFrac

Multiply fractions

tan a = \frac{3 \cdot 5}{5 \cdot 4}

CancelCommonFac

Cancel out common factors

tan a = \frac{3 \cdot 5}{5 \cdot 4}

SimpQuot

Simplify quotient

tan a = \frac{3}{4}

Let's now find the value of

cos b .

We will substitute the given value

sin b = - \frac{1 5}{1 7}

into the same identity as previously used.

sin^{2} b + cos^{2} b = 1

Substitute

$sin b = - \frac{1 5}{1 7}$

(- \frac{1 5}{1 7})^{2} + cos^{2} b = 1

▼

Solve for

cos b

NegBaseToPosPow

$(- a)^{2} = a^{2}$

(\frac{1 5}{1 7})^{2} + cos^{2} b = 1

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

\frac{2 2 5}{2 8 9} + cos^{2} b = 1

SubEqn

$LHS - \frac{2 2 5}{2 8 9} = RHS - \frac{2 2 5}{2 8 9}$

cos^{2} b = 1 - \frac{2 2 5}{2 8 9}

OneToFrac

Rewrite $1$ as $\frac{2 8 9}{2 8 9}$

cos^{2} b = \frac{2 8 9}{2 8 9} - \frac{2 2 5}{2 8 9}

SubFrac

Subtract fractions

cos^{2} b = \frac{6 4}{2 8 9}

SqrtEqn

$LHS = RHS$

cos b = \pm \frac{6 4}{2 8 9}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

cos b = \pm \frac{8}{1 7}

Let's now determine the sign of

cos b .

We are told that

b

is greater than

\frac{3 π}{2}

and less than

2 π .

Therefore, its terminal side is located in the fourth quadrant.

If the terminal side of an angle in standard position is in the fourth quadrant, then its cosine is positive. Therefore, we have that

cos b = \frac{8}{1 7} .

With this in mind, we can evaluate

tan b .

tan b = \frac{sin b}{cos b}

SubstituteII

$sin b = - \frac{1 5}{1 7}$ , $cos b = \frac{8}{1 7}$

tan b = - \frac{1 5}{1 7} / \frac{8}{1 7}

▼

Solve for

tan b

DivFracByFracSL

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

tan b = \frac{- 1 5}{1 7} \cdot \frac{1 7}{8}

MultFrac

Multiply fractions

tan b = \frac{- 1 5 \cdot 1 7}{1 7 \cdot 8}

CancelCommonFac

Cancel out common factors

tan b = \frac{- 1 5 \cdot 1 7}{1 7 \cdot 8}

SimpQuot

Simplify quotient

tan b = \frac{- 1 5}{8}

MoveNegDenomToFrac

Put minus sign in front of fraction

tan b = - \frac{1 5}{8}

Now we have all the information we need to calculate

tan (a - b) .

tan (a - b) = \frac{tan a - tan b}{1 + tan a tan b}

SubstituteII

$tan a = \frac{3}{4}$ , $tan b = - \frac{1 5}{8}$

tan (a - b) = \frac{\frac{3}{4} - ( - \frac{1 5}{8} )}{1 + \frac{3}{4} ( - \frac{1 5}{8} )}

▼

Evaluate right-hand side

SubNeg

$a - (- b) = a + b$

tan (a - b) = \frac{\frac{3}{4} + \frac{1 5}{8}}{1 + \frac{3}{4} ( - \frac{1 5}{8} )}

MultPosNeg

$a (- b) = - a \cdot b$

tan (a - b) = \frac{\frac{3}{4} + \frac{1 5}{8}}{1 + ( - \frac{3}{4} ( \frac{1 5}{8} ) )}

AddNeg

$a + (- b) = a - b$

tan (a - b) = \frac{\frac{3}{4} + \frac{1 5}{8}}{1 - \frac{3}{4} ( \frac{1 5}{8} )}

MultFrac

Multiply fractions

tan (a - b) = \frac{\frac{3}{4} + \frac{1 5}{8}}{1 - \frac{4 5}{3 2}}

OneToFrac

Rewrite $1$ as $\frac{3 2}{3 2}$

tan (a - b) = \frac{\frac{3}{4} + \frac{1 5}{8}}{\frac{3 2}{3 2} - \frac{4 5}{3 2}}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

tan (a - b) = \frac{\frac{6}{8} + \frac{1 5}{8}}{\frac{3 2}{3 2} - \frac{4 5}{3 2}}

AddSubFrac

Add and subtract fractions

tan (a - b) = \frac{\frac{2 1}{8}}{\frac{- 1 3}{3 2}}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 4}{b \cdot 4}$

tan (a - b) = \frac{\frac{2 1}{3 2}}{\frac{- 1 3}{3 2}}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 3 2}{b \cdot 3 2}$

tan (a - b) = \frac{2 1}{- 1 3}

MoveNegDenomToFrac

Put minus sign in front of fraction

tan (a - b) = - \frac{2 1}{1 3}