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Find the exact value of the trigonometric function given that sin u = -5/13 and cos v = -20/29. (Both u and v are in Quadrant III.)tan(u − v)

Question

Find the exact value of the trigonometric function given that sinu=513 \sin u = -\frac{5}{13} and cosv=2029 \cos v = -\frac{20}{29} . (Both u u and v v are in Quadrant III.) tan(uv) \tan(u − v)

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Solution

To find the exact value of the trigonometric function tan(u - v), we need to use the formula for the tangent of the difference of two angles, which is:

tan(u - v) = (tan(u) - tan(v)) / (1 + tan(u)tan(v))

First, we need to find the values of tan(u) and tan(v).

Since sin(u) = -5/13 and u is in Quadrant III, where both sine and cosine are negative, cos(u) can be found using the Pythagorean identity sin^2(u) + cos^2(u) = 1. Solving for cos(u), we get cos(u) = -sqrt(1 - sin^2(u)) = -sqrt(1 - (-5/13)^2) = -12/13. Therefore, tan(u) = sin(u) / cos(u) = (-5/13) / (-12/13) = 5/12.

Similarly, since cos(v) = -20/29 and v is in Quadrant III, sin(v) can be found using the Pythagorean identity sin^2(v) + cos^2(v) = 1. Solving for sin(v), we get sin(v) = -sqrt(1 - cos^2(v)) = -sqrt(1 - (-20/29)^2) = -21/29. Therefore, tan(v) = sin(v) / cos(v) = (-21/29) / (-20/29) = 21/20.

Substituting these values into the formula for tan(u - v), we get:

tan(u - v) = (5/12 - 21/20) / (1 + 5/12 * 21/20) = (-17/60) / (1 + 7/24) = -17/60 / 31/24 = -17/60 * 24/31 = -68/155.

So, the exact value of tan(u - v) is -68/155.

This problem has been solved

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