If $\cos A=\frac{7}{25}$ and $\tan B=\frac{3}{4}$ 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{7}{25}

and

\tan B=\frac{3}{4}

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 the sum of two angles: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ . First, we need to find $\tan A$ . Since we know $\cos A = \frac{7}{25}$ and $A$ is in Quadrant I, we can find $\sin A$ using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ . $\sin^2 A = 1 - \cos^2 A$ $\sin^2 A = 1 - \left(\frac{7}{25}\right)^2$ $\sin^2 A = 1 - \frac{49}{625}$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $0$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $1$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $2$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $3$ Now, $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $4$ .
Substitute values into identity: Now we have $\tan A = \frac{24}{7}$ and $\tan B = \frac{3}{4}$ . We can substitute these values into the identity for $\tan(A+B)$ . $\tan(A+B) = \frac{(\tan A + \tan B)}{(1 - \tan A \cdot \tan B)}$ $\tan(A+B) = \frac{(\frac{24}{7} + \frac{3}{4})}{(1 - (\frac{24}{7} \cdot \frac{3}{4}))}$
Simplify numerator and denominator: Simplify the numerator and the denominator separately. $\newline$ For the numerator: $\newline$ $\tan A + \tan B = \frac{24}{7} + \frac{3}{4}$ $\newline$ To add these fractions, find a common denominator, which is $28$ . $\newline$ $\tan A + \tan B = \frac{24 \times 4}{7 \times 4} + \frac{3 \times 7}{4 \times 7}$ $\newline$ $\tan A + \tan B = \frac{96}{28} + \frac{21}{28}$ $\newline$ $\tan A + \tan B = \frac{96 + 21}{28}$ $\newline$ $\tan A + \tan B = \frac{117}{28}$ $\newline$ For the denominator: $\newline$ $1 - \tan A \times \tan B = 1 - \left(\frac{24}{7} \times \frac{3}{4}\right)$ $\newline$ $1 - \tan A \times \tan B = 1 - \left(\frac{72}{28}\right)$ $\newline$ $1 - \tan A \times \tan B = \frac{28}{28} - \frac{72}{28}$ $\newline$ $1 - \tan A \times \tan B = \frac{-44}{28}$
Divide to find $\tan(A+B)$ : Now, divide the numerator by the denominator to find $\tan(A+B)$ . $\tan(A+B) = \frac{117}{28} / \left(-\frac{44}{28}\right)$ When dividing by a fraction, multiply by the reciprocal of the denominator. $\tan(A+B) = \frac{117}{28} \times \left(\frac{28}{-44}\right)$ The $28$ s cancel out. $\tan(A+B) = \frac{117}{-44}$ Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is $1$ in this case. $\tan(A+B) = -\frac{117}{44}$