If $\sin A=\frac{40}{41}$ and $\cos B=\frac{20}{29}$ and angles A and B are in Quadrant I, find the value of $\tan (A-B)$ . $\newline$ Answer:

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Q. If

\sin A=\frac{40}{41}

and

\cos B=\frac{20}{29}

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

\tan (A-B)

\newline

Answer:

Find $\cos(A)$ : Use the Pythagorean identity to find $\cos(A)$ . Since $\sin(A) = \frac{40}{41}$ and $A$ is in Quadrant I, we can use the Pythagorean identity $\sin^2(A) + \cos^2(A) = 1$ to find $\cos(A)$ . $\cos^2(A) = 1 - \sin^2(A)$ $\cos^2(A) = 1 - \left(\frac{40}{41}\right)^2$ $\cos^2(A) = 1 - \frac{1600}{1681}$ $\cos^2(A) = \frac{1681}{1681} - \frac{1600}{1681}$ $\cos(A)$ $0$ $\cos(A)$ $1$ $\cos(A)$ $2$
Find $\sin(B)$ : Use the Pythagorean identity to find $\sin(B)$ . $\newline$ Since $\cos(B) = \frac{20}{29}$ and $B$ is in Quadrant I, we can use the Pythagorean identity $\sin^2(B) + \cos^2(B) = 1$ to find $\sin(B)$ . $\newline$ $\sin^2(B) = 1 - \cos^2(B)$ $\newline$ $\sin^2(B) = 1 - \left(\frac{20}{29}\right)^2$ $\newline$ $\sin^2(B) = 1 - \frac{400}{841}$ $\newline$ $\sin^2(B) = \frac{841}{841} - \frac{400}{841}$ $\newline$ $\sin(B)$ $0$ $\newline$ $\sin(B)$ $1$ $\newline$ $\sin(B)$ $2$
Find $\tan(A-B)$ : Use the angle subtraction formula for tangent to find $\tan(A-B)$ . The formula for $\tan(A-B)$ is $\frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}$ . We already have $\sin(A)$ and $\cos(A)$ , and $\sin(B)$ and $\cos(B)$ , so we can find $\tan(A)$ and $\tan(B)$ . $\tan(A-B)$ $0$ $\tan(A-B)$ $1$ $\tan(A-B)$ $2$ $\tan(A-B)$ $3$ $\tan(A-B)$ $4$ $\tan(A-B)$ $5$ Now we can find $\tan(A-B)$ . $\tan(A-B)$ $7$ $\tan(A-B)$ $8$
Find $\tan(A-B)$ : Use the angle subtraction formula for tangent to find $\tan(A-B)$ . The formula for $\tan(A-B)$ is $\frac{\tan(A) - \tan(B)}{1 + \tan(A)\tan(B)}$ . We already have $\sin(A)$ and $\cos(A)$ , and $\sin(B)$ and $\cos(B)$ , so we can find $\tan(A)$ and $\tan(B)$ . $\tan(A-B)$ $0$ $\tan(A-B)$ $1$ $\tan(A-B)$ $2$ $\tan(A-B)$ $3$ $\tan(A-B)$ $4$ $\tan(A-B)$ $5$ Now we can find $\tan(A-B)$ . $\tan(A-B)$ $7$ $\tan(A-B)$ $8$ Simplify the expression for $\tan(A-B)$ . $\tan(A-B)$ $8$ To subtract the fractions, find a common denominator. Common denominator for $\tan(A-B)$ $1$ and $\tan(A-B)$ $2$ is $\tan(A-B)$ $3$ . $\tan(A-B)$ $4$ $\tan(A-B)$ $5$ $\tan(A-B)$ $6$ $\tan(A-B)$ $7$ Now simplify the expression. $\tan(A-B)$ $8$ $\tan(A-B)$ $9$