If $\cos A=\frac{28}{53}$ and $\tan B=\frac{5}{12}$ and angles A and B are in Quadrant I, find the value of $\tan (A+B)$ . $\newline$ Answer:

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Q. If

\cos A=\frac{28}{53}

and

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

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

\tan (A+B)

\newline

Answer:

Find $\tan A$ : Use the identity for the tangent of a sum of two angles: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ . First, we need to find $\tan A$ using the given $\cos A = \frac{28}{53}$ . Since $\cos A = \frac{\text{adjacent}}{\text{hypotenuse}}$ , we can find the opposite side using the Pythagorean theorem: $\text{opposite}^2 = \text{hypotenuse}^2 - \text{adjacent}^2$ .
Calculate opposite side: Calculate the opposite side for angle $A$ : $\text{opposite}^2 = 53^2 - 28^2.$ $\text{opposite}^2 = 2809 - 784.$ $\text{opposite}^2 = 2025.$ $\text{opposite} = \sqrt{2025}.$ $\text{opposite} = 45.$
Find $\tan B$ : Now we can find $\tan A$ : $\tan A = \frac{\text{opposite}}{\text{adjacent}}.$ $\newline$ $\tan A = \frac{45}{28}.$
Apply $\tan(A+B)$ identity: We already have $\tan B$ given as $\frac{5}{12}$ . $\newline$ Now we can use the identity for $\tan(A+B)$ : $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ . $\newline$ $\tan(A+B) = \frac{\frac{45}{28} + \frac{5}{12}}{1 - \left(\frac{45}{28} \cdot \frac{5}{12}\right)}$ .
Add fractions: Find a common denominator for adding the fractions $\frac{45}{28}$ and $\frac{5}{12}$ . The common denominator is $84$ . Convert the fractions: $\left(\frac{45}{28}\right) \times \left(\frac{3}{3}\right) = \frac{135}{84}$ and $\left(\frac{5}{12}\right) \times \left(\frac{7}{7}\right) = \frac{35}{84}$ . Now add the fractions: $\frac{135}{84} + \frac{35}{84} = \frac{170}{84}$ .
Simplify fraction: Simplify the fraction $\frac{170}{84}$ by dividing both numerator and denominator by $2$ . $\newline$ $\frac{170}{84} = \frac{85}{42}$ .
Calculate denominator: Now calculate the denominator of the $\tan(A+B)$ formula: $1 - \left(\frac{45}{28} \times \frac{5}{12}\right)$ . First, multiply the fractions: $\left(\frac{45}{28}\right) \times \left(\frac{5}{12}\right) = \frac{225}{336}$ . Simplify the fraction $\frac{225}{336}$ by dividing both numerator and denominator by $3$ . $\frac{225}{336} = \frac{75}{112}.$
Subtract fraction from $1$ : Now subtract the fraction from $1$ : $1 - \frac{75}{112}$ . Convert $1$ to a fraction with the same denominator: $\frac{112}{112} - \frac{75}{112} = \frac{37}{112}$ .
Calculate $\tan(A+B)$ formula: Now we have both the numerator and the denominator for the $\tan(A+B)$ formula. $\newline$ $\tan(A+B) = \frac{85}{42} / \frac{37}{112}$ . $\newline$ To divide by a fraction, multiply by its reciprocal: $\frac{85}{42} \times \frac{112}{37}$ .
Multiply fractions: Multiply the fractions: $(85 \times 112) / (42 \times 37)$ . $85 \times 112 = 9520$ and $42 \times 37 = 1554$ . $\tan(A+B) = \frac{9520}{1554}$ .
Simplify fraction: Simplify the fraction $\frac{9520}{1554}$ by finding the greatest common divisor (GCD) and dividing both numerator and denominator by it. $\newline$ The GCD of $9520$ and $1554$ is $2$ . $\newline$ $\frac{9520}{2} = 4760$ and $\frac{1554}{2} = 777$ . $\newline$ $\tan(A+B) = \frac{4760}{777}$ .
Simplify further: Simplify the fraction $\frac{4760}{777}$ further if possible. $\newline$ The GCD of $4760$ and $777$ is $1$ , so the fraction is already in its simplest form. $\newline$ $\tan(A+B) = \frac{4760}{777}$ .