If $\cos A=\frac{21}{29}$ and $\tan B=\frac{9}{40}$ 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{21}{29}

and

\tan B=\frac{9}{40}

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

\tan (A-B)

\newline

Answer:

Find $\tan A$ : Use the formula for $\tan(A-B)$ , which is $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$ . 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 $\sin^2 A + \cos^2 A = 1$ . $\sin^2 A = 1 - \cos^2 A$ $\sin^2 A = 1 - \left(\frac{21}{29}\right)^2$ $\tan(A-B)$ $0$ $\tan(A-B)$ $1$ $\tan(A-B)$ $2$ $\tan(A-B)$ $3$ $\tan(A-B)$ $4$ Now, $\tan(A-B)$ $5$ $\tan(A-B)$ $6$ $\tan(A-B)$ $7$
Substitute values into formula: Now we have $\tan A = \frac{20}{21}$ and $\tan B = \frac{9}{40}$ . We can substitute these values into the $\tan(A-B)$ formula. $\newline$ $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$ $\newline$ $\tan(A-B) = \frac{\frac{20}{21} - \frac{9}{40}}{1 + (\frac{20}{21} \cdot \frac{9}{40})}$
Simplify numerator and denominator: Simplify the numerator and the denominator separately. $\newline$ For the numerator: $\newline$ $\tan A - \tan B = \frac{20}{21} - \frac{9}{40}$ $\newline$ To subtract these fractions, find a common denominator, which is $21\times40 = 840$ . $\newline$ $(\frac{20}{21} - \frac{9}{40}) = (\frac{20\times40}{21\times40}) - (\frac{9\times21}{40\times21})$ $\newline$ $(\frac{20}{21} - \frac{9}{40}) = (\frac{800}{840}) - (\frac{189}{840})$ $\newline$ $(\frac{20}{21} - \frac{9}{40}) = \frac{800 - 189}{840}$ $\newline$ $(\frac{20}{21} - \frac{9}{40}) = \frac{611}{840}$ $\newline$ For the denominator: $\newline$ $1 + \tan A \times \tan B = 1 + (\frac{20}{21} \times \frac{9}{40})$ $\newline$ $1 + \tan A \times \tan B = 1 + (\frac{180}{840})$ $\newline$ $1 + \tan A \times \tan B = \frac{840}{840} + \frac{180}{840}$ $\newline$ $1 + \tan A \times \tan B = \frac{1020}{840}$
Divide to find $\tan(A-B)$ : Now, divide the numerator by the denominator to find $\tan(A-B)$ . $\tan(A-B) = \frac{611}{840} / \frac{1020}{840}$ When dividing by a fraction, multiply by the reciprocal of the divisor. $\tan(A-B) = \frac{611}{840} \times \frac{840}{1020}$ Simplify by canceling out the common factor of $840$ . $\tan(A-B) = \frac{611}{1020}$ Reduce the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is $1$ . $\tan(A-B) = \frac{611}{1020}$