If tan A=(35)/(12) and cos B=(5)/(13) and angles A and B are in Quadrant I, find the value of tan(A+B). Answer:

Apply Tangent Addition Formula: Use the tangent addition formula: tan(A+B) = (tan A + tan B) / (1 - tan A * tan B). First, we need to find tan B. Since cos B = 5/13 and B is in Quadrant I, 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 - (5/13)^2 sin^2 B = 1 - 25/169 sin^2 B = (169/169) - (25/169) sin^2 B = 144/169 sin B = √(144/169) sin B = 12/13Find sin B: Now that we have sin B, we can find tan B using the definition tan B = sin B / cos B. tan B = (12/13) / (5/13) tan B = (12/13) * (13/5) tan B = 12/5Find tan B: Now we have both tan A and tan B, so we can use the tangent addition formula. tan(A+B) = (tan A + tan B) / (1 - tan A * tan B) tan(A+B) = ((35/12) + (12/5)) / (1 - (35/12) * (12/5))Use Tangent Addition Formula: Simplify the numerator and the denominator separately. Numerator: (35/12) + (12/5) = (35*5 + 12*12) / (12*5) = (175 + 144) / 60 = 319/60 Denominator: 1 - (35/12) * (12/5) = 1 - (35*12) / (12*5) = 1 - (35*12) / (60) = 1 - 7 = -6Simplify Numerator and Denominator: Now, divide the numerator by the denominator to find tan(A+B). tan(A+B) = (319/60) / (-6) tan(A+B) = 319/60 * -1/6 tan(A+B) = -319/360

Home

Math Problems

Algebra 2

Find trigonometric ratios using multiple identities

Full solution

Q. If

\tan A=\frac{35}{12}

and

\cos B=\frac{5}{13}

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

\tan (A+B)

\newline

Answer:

Apply Tangent Addition Formula: Use the tangent addition formula: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ . First, we need to find $\tan B$ . Since $\cos B = \frac{5}{13}$ and $B$ is in Quadrant I, we can find $\sin B$ using the Pythagorean identity $\sin^2 B + \cos^2 B = 1$ . $\sin^2 B = 1 - \cos^2 B$ $\sin^2 B = 1 - (\frac{5}{13})^2$ $\sin^2 B = 1 - \frac{25}{169}$ $\sin^2 B = \frac{169}{169} - \frac{25}{169}$ $\tan B$ $0$ $\tan B$ $1$ $\tan B$ $2$
Find $\sin B$ : Now that we have $\sin B$ , we can find $\tan B$ using the definition $\tan B = \frac{\sin B}{\cos B}$ . $\newline$ $\tan B = \frac{12/13}{5/13}$ $\newline$ \tan B = \frac{12/13} \times \frac{13/5} $\newline$ $\tan B = \frac{12}{5}$
Find $\tan B$ : Now we have both $\tan A$ and $\tan B$ , so we can use the tangent addition formula. $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $\tan(A+B) = \frac{(\frac{35}{12}) + (\frac{12}{5})}{1 - (\frac{35}{12}) \cdot (\frac{12}{5})}$
Use Tangent Addition Formula: Simplify the numerator and the denominator separately. $\newline$ Numerator: $(\frac{35}{12}) + (\frac{12}{5}) = (\frac{35\times 5 + 12\times 12}{12\times 5}) = (\frac{175 + 144}{60}) = \frac{319}{60}$ $\newline$ Denominator: $1 - (\frac{35}{12}) \times (\frac{12}{5}) = 1 - (\frac{35\times 12}{12\times 5}) = 1 - (\frac{35\times 12}{60}) = 1 - 7 = -6$
Simplify Numerator and Denominator: Now, divide the numerator by the denominator to find $\tan(A+B)$ . $\newline$ $\tan(A+B) = \frac{319}{60} / (-6)$ $\newline$ $\tan(A+B) = \frac{319}{60} \times -\frac{1}{6}$ $\newline$ $\tan(A+B) = -\frac{319}{360}$