If $\tan A=\frac{28}{45}$ and $\cos B=\frac{4}{5}$ and angles A and B are in Quadrant I, find the value of $\tan (A-B)$ . $\newline$ Answer:

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Q. If

\tan A=\frac{28}{45}

and

\cos B=\frac{4}{5}

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

\tan (A-B)

\newline

Answer:

Given $\tan(A)$ : We know that $\tan(A) = \frac{28}{45}$ . To find $\tan(A-B)$ , we can use the formula for the tangent of the difference of two angles: $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$ . First, we need to find $\tan B$ , which requires finding $\sin B$ since we already have $\cos B$ .
Find $\sin B$ : We have $\cos B = \frac{4}{5}$ . Since $B$ is in Quadrant I, where all trigonometric functions are positive, we can find $\sin B$ using the Pythagorean identity $\sin^2 B + \cos^2 B = 1$ . Let's calculate $\sin B$ . $\newline$ $\sin^2 B = 1 - \cos^2 B$ $\newline$ $\sin^2 B = 1 - \left(\frac{4}{5}\right)^2$ $\newline$ $\sin^2 B = 1 - \frac{16}{25}$ $\newline$ $\sin^2 B = \frac{25}{25} - \frac{16}{25}$ $\newline$ $\cos B = \frac{4}{5}$ $0$ $\newline$ $\cos B = \frac{4}{5}$ $1$ $\newline$ $\cos B = \frac{4}{5}$ $2$
Find $\tan B$ : Now that we have $\sin B$ , we can find $\tan B$ using the definition $\tan B = \frac{\sin B}{\cos B}$ .
$\tan B = \frac{3/5}{4/5}$
$\tan B = \frac{3}{5} \times \frac{5}{4}$
$\tan B = \frac{3}{4}$
Use formula for $\tan(A-B)$ : With $\tan A = \frac{28}{45}$ and $\tan B = \frac{3}{4}$ , we can now use the formula for $\tan(A-B)$ . $\newline$ $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$ $\newline$ $\tan(A-B) = \frac{\frac{28}{45} - \frac{3}{4}}{1 + (\frac{28}{45} \cdot \frac{3}{4})}$
Simplify numerator: Let's simplify the numerator and the denominator separately. $\newline$ First, the numerator: $\newline$ $\tan A - \tan B = \frac{28}{45} - \frac{3}{4}$ $\newline$ To subtract these fractions, we need a common denominator, which is $180$ . $\newline$ $\tan A - \tan B = \frac{28 \times 4}{45 \times 4} - \frac{3 \times 45}{4 \times 45}$ $\newline$ $\tan A - \tan B = \frac{112}{180} - \frac{135}{180}$ $\newline$ $\tan A - \tan B = \frac{112 - 135}{180}$ $\newline$ $\tan A - \tan B = -\frac{23}{180}$
Simplify denominator: Now, let's simplify the denominator: $\newline$ $1 + \tan A \cdot \tan B = 1 + \left(\frac{28}{45} \cdot \frac{3}{4}\right)$ $\newline$ $1 + \tan A \cdot \tan B = 1 + \left(\frac{28 \cdot 3}{45 \cdot 4}\right)$ $\newline$ $1 + \tan A \cdot \tan B = 1 + \frac{84}{180}$ $\newline$ $1 + \tan A \cdot \tan B = 1 + \frac{7}{15}$ $\newline$ To add $1$ to $\frac{7}{15}$ , we need to express $1$ as $\frac{15}{15}$ . $\newline$ $1 + \tan A \cdot \tan B = \frac{15}{15} + \frac{7}{15}$ $\newline$ $1 + \tan A \cdot \tan B = \frac{(15 + 7)}{15}$ $\newline$ $1 + \tan A \cdot \tan B = 1 + \left(\frac{28 \cdot 3}{45 \cdot 4}\right)$ $0$
Divide numerator by denominator: Now we can divide the numerator by the denominator to find $\tan(A-B)$ . $\tan(A-B) = \frac{-23}{180} / \frac{22}{15}$ To divide by a fraction, we multiply by its reciprocal. $\tan(A-B) = \frac{-23}{180} \times \frac{15}{22}$
Simplify multiplication: Let's simplify the multiplication by canceling out common factors. $\newline$ $\tan(A-B) = (-23 \times 15) / (180 \times 22)$ $\newline$ $\tan(A-B) = (-345) / (3960)$ $\newline$ Both numerator and denominator are divisible by $15$ . $\newline$ $\tan(A-B) = (-23) / (264)$
Further simplify: We can further simplify by dividing both numerator and denominator by $23$ . $\newline$ $\tan(A-B) = \frac{-1}{264/23}$ $\newline$ $\tan(A-B) = \frac{-1}{11.4782608696}$