If $\cos A=\frac{21}{29}$ 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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Find trigonometric ratios using multiple identities

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Q. If

\cos A=\frac{21}{29}

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 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 $\cos A = \frac{21}{29}$ and $A$ is in Quadrant I, we can find $\sin A$ using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ . $\sin^2 A = 1 - \cos^2 A$ $\sin^2 A = 1 - \left(\frac{21}{29}\right)^2$ $\sin^2 A = 1 - \frac{441}{841}$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $0$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $1$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $2$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $3$ Now, $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $4$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $5$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $6$
Substitute values into identity: Now we have $\tan A = \frac{20}{21}$ and $\tan B = \frac{5}{12}$ . We can substitute these values into the identity for $\tan(A+B)$ . $\newline$ $\tan(A+B) = \frac{(\tan A + \tan B)}{(1 - \tan A \cdot \tan B)}$ $\newline$ $\tan(A+B) = \frac{(\frac{20}{21} + \frac{5}{12})}{(1 - (\frac{20}{21} \cdot \frac{5}{12}))}$
Simplify numerator and denominator: Simplify the numerator and the denominator separately. $\newline$ For the numerator: $\newline$ $\tan A + \tan B = \frac{20}{21} + \frac{5}{12}$ $\newline$ To add these fractions, find a common denominator, which is $21 \times 12 = 252$ . $\newline$ $\left(\frac{20}{21}\right) \times \left(\frac{12}{12}\right) + \left(\frac{5}{12}\right) \times \left(\frac{21}{21}\right) = \frac{240}{252} + \frac{105}{252}$ $\newline$ $= \frac{240 + 105}{252}$ $\newline$ $= \frac{345}{252}$ $\newline$ For the denominator: $\newline$ $1 - \left(\frac{20}{21} \times \frac{5}{12}\right) = 1 - \frac{100}{252}$ $\newline$ To subtract these fractions, express $1$ as $\frac{252}{252}$ . $\newline$ $\left(\frac{252}{252}\right) - \frac{100}{252} = \frac{252 - 100}{252}$ $\newline$ $= \frac{152}{252}$
Divide to find $\tan(A+B)$ : Now we can divide the numerator by the denominator to find $\tan(A+B)$ . $\tan(A+B) = \frac{345}{252} / \frac{152}{252}$ When dividing by a fraction, multiply by the reciprocal of the divisor. $\tan(A+B) = \frac{345}{252} \times \frac{252}{152}$ The $252$ in the numerator and denominator cancel out. $\tan(A+B) = \frac{345}{152}$
Simplify final fraction: Simplify the fraction $\frac{345}{152}$ if possible. $\newline$ $345$ and $152$ have no common factors other than $1$ , so the fraction is already in its simplest form.