Full solution

Q. If

\cos A=\frac{60}{61}

and

\tan B=\frac{8}{15}

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

\tan (A+B)

\newline

Answer:

Use tangent identity: Use the identity for the tangent of a sum of two angles. $\newline$ The identity is $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ . $\newline$ We know $\tan B$ , but we need to find $\tan A$ .
Find $\tan A$ : Find $\tan A$ using the given value of $\cos A$ and the Pythagorean identity. $\newline$ Since $\cos A = \frac{\text{adjacent}}{\text{hypotenuse}}$ , we have $\text{adjacent} = 60$ and $\text{hypotenuse} = 61$ . $\newline$ To find the opposite side, we use the Pythagorean theorem: $\text{opposite}^2 = \text{hypotenuse}^2 - \text{adjacent}^2$ .
Calculate opposite side: Calculate the opposite side for angle $A$ . $\newline$ $\text{opposite}^2 = 61^2 - 60^2 = 3721 - 3600 = 121$ . $\newline$ Taking the square root gives us $\text{opposite} = 11$ .
Calculate $\tan A$ : Calculate $\tan A$ using the opposite and adjacent sides. $\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{11}{60}$ .
Substitute into identity: Substitute $\tan A$ and $\tan B$ into the identity for $\tan(A+B)$ . $\newline$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} = \frac{\frac{11}{60} + \frac{8}{15}}{1 - (\frac{11}{60} \cdot \frac{8}{15}))$.
Simplify expression: Simplify the expression for $\tan(A+B)$. First, find a common denominator for the sum in the numerator: $60 \times 15 = 900$. $\tan(A+B) = \frac{(11 \times 15) + (8 \times 60)}{900 - (11 \times 8)} = \frac{(165 + 480)}{(900 - 88)}$.
Reduce fraction: Continue simplifying the expression. $\tan(A+B) = \frac{645}{900 - 88} = \frac{645}{812}$.
Reduce fraction: Continue simplifying the expression. $\tan(A+B) = \frac{645}{900 - 88} = \frac{645}{812}$. Reduce the fraction to its simplest form if possible. The fraction $\frac{645}{812}$ cannot be simplified further as they do not have common factors other than $1$.