If $\tan A=\frac{35}{12}$ and $\cos B=\frac{21}{29}$ 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{35}{12}

and

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

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

\tan (A+B)

\newline

Answer:

Use $\tan(A+B)$ Formula: Use the formula for $\tan(A+B)$ , which is $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ . We know $\tan A = \frac{35}{12}$ , but we need to find $\tan B$ using the given $\cos B = \frac{21}{29}$ .
Find $\tan B$ : To find $\tan B$ , we first need to find $\sin B$ . Since $\cos^2 B + \sin^2 B = 1$ , we can solve for $\sin B$ . $\newline$ $\sin^2 B = 1 - \cos^2 B$ $\newline$ $\sin^2 B = 1 - (\frac{21}{29})^2$ $\newline$ $\sin^2 B = 1 - (\frac{441}{841})$ $\newline$ $\sin^2 B = (841 - 441) / 841$ $\newline$ $\sin^2 B = \frac{400}{841}$ $\newline$ $\tan B$ $0$ $\newline$ $\tan B$ $1$ , since $\tan B$ $2$ is in Quadrant I, $\sin B$ is positive.
Calculate $\tan B$ : Now that we have $\sin B$ , we can find $\tan B$ .
$\tan B = \frac{\sin B}{\cos B}$
$\tan B = \frac{20/29}{21/29}$
$\tan B = \frac{20}{21}$
Substitute $\tan A$ and $\tan B$ : Substitute $\tan A$ and $\tan B$ into the $\tan(A+B)$ formula. $\newline$ $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$ $\newline$ $\tan(A+B) = \frac{\frac{35}{12} + \frac{20}{21}}{1 - \left(\frac{35}{12} \cdot \frac{20}{21}\right)}$
Simplify numerator and denominator: Simplify the numerator and the denominator separately. $\newline$ First, find a common denominator for the numerator. $\newline$ The common denominator for $12$ and $21$ is $84$ . $\newline$ $\tan(A+B) = \left(\frac{35 \times 7}{12 \times 7} + \frac{20 \times 4}{21 \times 4}\right) / \left(1 - \frac{35}{12} \times \frac{20}{21}\right)$ $\newline$ $\tan(A+B) = \frac{245}{84} + \frac{80}{84} / \left(1 - \frac{700}{252}\right)$
Continue simplifying: Continue simplifying the numerator and the denominator. $\newline$ $\tan(A+B) = \frac{325}{84} / \left(1 - \frac{700}{252}\right)$ $\newline$ Now, simplify the denominator: $\newline$ $1 - \frac{700}{252} = \frac{252}{252} - \frac{700}{252}$ $\newline$ $1 - \frac{700}{252} = \frac{252 - 700}{252}$ $\newline$ $1 - \frac{700}{252} = \frac{-448}{252}$ $\newline$ $1 - \frac{700}{252} = \frac{-56}{63}$ (simplifying the fraction by dividing both numerator and denominator by $4$ )
Divide numerator by denominator: Now, divide the numerator by the denominator. $\newline$ $\tan(A+B) = \frac{325}{84} / \frac{-56}{63}$ $\newline$ To divide by a fraction, multiply by its reciprocal. $\newline$ $\tan(A+B) = \frac{325}{84} \times \frac{63}{-56}$
Simplify multiplication: Simplify the multiplication. $\newline$ $\tan(A+B) = \frac{325 \times 63}{84 \times -56}$ $\newline$ $\tan(A+B) = \frac{20475}{-4704}$ $\newline$ $\tan(A+B) = -\frac{20475}{4704}$ (keeping the negative sign in the numerator)
Check for further simplification: Check if the fraction can be simplified further. $\newline$ Both $20475$ and $4704$ are divisible by $3$ . $\newline$ $\tan(A+B) = \frac{-6825}{1568}$ $\newline$ This fraction cannot be simplified further.