SOLUTION: Given sin(A) = 5/13 with 90 degrees < A < 180 degrees and cos(B) = 3/5 with 270 degrees < B < 360 degrees, which expression gives the value of tan(A-B)? a. 33/56 b. -33/16 c.

Algebra ->  Trigonometry-basics -> SOLUTION: Given sin(A) = 5/13 with 90 degrees < A < 180 degrees and cos(B) = 3/5 with 270 degrees < B < 360 degrees, which expression gives the value of tan(A-B)? a. 33/56 b. -33/16 c.       Log On


   

Question 958927: Given sin(A) = 5/13 with 90 degrees < A < 180 degrees and cos(B) = 3/5 with 270 degrees < B < 360 degrees, which expression gives the value of tan(A-B)?
a. 33/56
b. -33/16
c. -33/56
d. 33/16
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
To find tan(A-B) we will probably need to use the identity:
tan%28A-B%29+=+%28tan%28A%29-tan%28B%29%29%2F%281%2Btan%28A%29%2Atan%28B%29%29

This identity requires what we have tan(A) and tan(B). Since we were given sin(A) and cos(B) (and the quadrants in which they terminate), we will use the reciprocal identity:
tan%28x%29+=+sin%28x%29%2Fcos%28x%29
to find tan(A) and tan(B).

The recirpocal identity requires that we have both the sin and cos of an angle. We have only sin(A) and cos(B). So we start the solution by finding cos(A) and sin(B)

We will break up finding cos(A) and sin(B) into two parts: using the quadrant to determine the sign and using the given sin(A) to find the ratio for cos(A) and using the given cos(B) to find the ratio for sin(B).

Since A terminates in the 2nd quadrant and since cos is negative in the 2nd quadrant, cos(A) will be negative. Since B terminates in the 4th quadrant and since sin is negative in the 4th quadrant, sin(B) will be negative.

For the ratio portion of the cos(A) we can use the Pythagorean identity:
cos%5E2%28x%29+=+1+-+sin%5E2%28x%29
Substituting in our given sin(A):
cos%5E2%28A%29+=+1+-+%285%2F13%29%5E2
Simplifying...
cos%5E2%28A%29+=+1+-+25%2F169
cos%5E2%28A%29+=+169%2F169+-+25%2F169
cos%5E2%28A%29+=+144%2F169
Now we find the square root of each side. From the fact that A terminates in the 2nd quadrant, we know to use the negative square root:
sqrt%28cos%5E2%28A%29%29+=+-sqrt%28144%2F169%29
Simplifying...
cos%28A%29+=+-12%2F13

Repeating this for sin(B):
sin%5E2%28x%29+=+1+-+cos%5E2%28x%29
Substituting in our given cos(B):
sin%5E2%28B%29+=+1+-+%283%2F5%29%5E2
Simplifying...
sin%5E2%28B%29+=+1+-+9%2F25
sin%5E2%28B%29+=+25%2F25+-+9%2F25
sin%5E2%28B%29+=+16%2F25
Now we find the square root of each side. From the fact that B terminates in the 4th quadrant, we know to use the negative square root:
sqrt%28sin%5E2%28B%29%29+=+-sqrt%2816%2F25%29
Simplifying...
sin%28B%29+=+-4%2F5

Now we're ready to find tan(A) and tan(B):



Now we're ready to find tan(A-B):