If $\tan A=\frac{12}{5}$ and $\cos B=\frac{8}{17}$ and angles A and B are in Quadrant I, find the value of $\tan (A-B)$ . $\newline$ Answer:

View step-by-step help

Home

Math Problems

Algebra 2

Find trigonometric ratios using multiple identities

Full solution

Q. If

\tan A=\frac{12}{5}

and

\cos B=\frac{8}{17}

and angles A and B are in Quadrant I, find the value of

\tan (A-B)

\newline

Answer:

Apply $\tan(A-B)$ formula: Use the formula for $\tan(A-B)$ , which is $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$ . We know $\tan A = \frac{12}{5}$ . We need to find $\tan B$ using the given $\cos B = \frac{8}{17}$ .
Find $\tan B$ : Since $\cos B = \frac{8}{17}$ , we can find $\sin B$ using the Pythagorean identity $\sin^2 B + \cos^2 B = 1$ .
$\sin^2 B = 1 - \cos^2 B$
$\sin^2 B = 1 - \left(\frac{8}{17}\right)^2$
$\sin^2 B = 1 - \frac{64}{289}$
$\sin^2 B = \frac{289 - 64}{289}$
$\sin^2 B = \frac{225}{289}$
$\sin B = \sqrt{\frac{225}{289}}$
$\cos B = \frac{8}{17}$ $0$
Calculate $\tan(A-B)$ : Now we can find $\tan B$ using the ratio $\tan B = \frac{\sin B}{\cos B}$ .
$\tan B = \frac{15}{17} / \frac{8}{17}$
$\tan B = \frac{15}{17} \cdot \frac{17}{8}$
$\tan B = \frac{15}{8}$
Simplify numerator and denominator: Substitute the values of $\tan A$ and $\tan B$ into the $\tan(A-B)$ formula. $\newline$ $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$ $\newline$ $\tan(A-B) = \frac{(\frac{12}{5}) - (\frac{15}{8})}{1 + (\frac{12}{5}) \cdot (\frac{15}{8})}$
Divide to find $\tan(A-B)$ : Simplify the numerator and the denominator separately. $\newline$ Numerator: $\frac{12}{5}$ - $\frac{15}{8}$ = $\frac{96}{40}$ - $\frac{75}{40}$ = $\frac{96 - 75}{40}$ = $\frac{21}{40}$ $\newline$ Denominator: $1 + \frac{12}{5} \times \frac{15}{8}$ = $1 + \frac{180}{40}$ = $1 + 4.5$ = $\frac{12}{5}$ $0$
Divide to find $\tan(A-B)$ : Simplify the numerator and the denominator separately. $\newline$ Numerator: $(12/5) - (15/8) = (96/40) - (75/40) = (96 - 75) / 40 = 21 / 40$ $\newline$ Denominator: $1 + (12/5) \times (15/8) = 1 + (180/40) = 1 + 4.5 = 5.5$ Now divide the numerator by the denominator to find $\tan(A-B)$ . $\newline$ $\tan(A-B) = (21 / 40) / (5.5)$ $\newline$ $\tan(A-B) = (21 / 40) \times (1 / 5.5)$ $\newline$ $\tan(A-B) = 21 / (40 \times 5.5)$ $\newline$ $\tan(A-B) = 21 / 220$ $\newline$ $\tan(A-B) = 3 / 40$