If $\cos A=\frac{9}{41}$ 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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Q. If

\cos A=\frac{9}{41}

and

\tan B=\frac{5}{12}

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

\tan (A-B)

\newline

Answer:

Use $\tan(A-B)$ Identity: Use the identity for $\tan(A-B)$ , which is $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$ . We need to find $\tan A$ and $\tan B$ to use this formula.
Find $\sin A$ : We know that $\cos A = \frac{9}{41}$ and $\tan B = \frac{5}{12}$ . Since $A$ is in Quadrant I, $\sin A$ will be positive. Use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ to find $\sin A$ . $\sin^2 A = 1 - \cos^2 A$ $\sin^2 A = 1 - \left(\frac{9}{41}\right)^2$ $\sin^2 A = 1 - \frac{81}{1681}$ $\cos A = \frac{9}{41}$ $0$ $\cos A = \frac{9}{41}$ $1$ $\cos A = \frac{9}{41}$ $2$ $\cos A = \frac{9}{41}$ $3$
Find $\tan A$ : Now, find $\tan A$ using the definition $\tan A = \frac{\sin A}{\cos A}$ . $\tan A = \frac{(40/41)}{(9/41)}$ $\tan A = \frac{40}{9}$
Substitute $\tan A$ and $\tan B$ : Substitute the values of $\tan A$ and $\tan B$ into the identity for $\tan(A-B)$ . $\newline$ $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$ $\newline$ $\tan(A-B) = \frac{\frac{40}{9} - \frac{5}{12}}{1 + \left(\frac{40}{9}\right) \cdot \left(\frac{5}{12}\right)}$
Simplify with Common Denominator: Simplify the expression by finding a common denominator for $\tan A$ and $\tan B$ . $\tan(A-B) = \left(\frac{40}{9} \times \frac{12}{12} - \frac{5}{12} \times \frac{9}{9}\right) / \left(1 + \frac{40}{9} \times \frac{5}{12}\right)$ $\tan(A-B) = \left(\frac{480}{108} - \frac{45}{108}\right) / \left(1 + \frac{200}{108}\right)$ $\tan(A-B) = \left(\frac{435}{108}\right) / \left(1 + \frac{200}{108}\right)$
Simplify Denominator: Simplify the denominator of the expression. $\newline$ $1 + \frac{200}{108} = \frac{108}{108} + \frac{200}{108}$ $\newline$ $1 + \frac{200}{108} = \frac{308}{108}$
Divide Numerator by Denominator: Now, divide the numerator by the denominator. $\newline$ $\tan(A-B) = \frac{435}{108} / \frac{308}{108}$ $\newline$ $\tan(A-B) = \frac{435}{108} \times \frac{108}{308}$
Simplify Expression: Simplify the expression by canceling out the common factor of $108$ . $\newline$ $\tan(A-B) = \frac{435}{308}$
Check for Further Simplification: Check if the fraction can be simplified further. $435$ and $308$ have no common factors other than $1$ , so the fraction is already in its simplest form.