Answer:
-1239/1961
Explanation:
You want cos(A+B) given that tan(A) = 45/28 and cos(B) = 12/37.
Tangent formulas
Here, we'll use the tangent relations ...
- tan(A+B) = (tan(A) +tan(B))/(1 -tan(A)tan(B))
- tan(x)² +1 = sec(x)² = 1/cos(x)²
Application
The tangent of angle B can be found from ...
tan(B)² = 1/cos(B)² -1
tan(B)² = 1/(12/37)² -1 = 1225/144
tan(B) = 35/12
Now the tangent of the angle sum is ...
tan(A+B) = (tan(A) +tan(B))/(1 -tan(A)tan(B))
= (45/28 +35/12)/(1 -(45/28)(35/12)) = (95/21)/(1 -75/16) = -1520/1239
Cosine
Note that the tangent of this sum of two first-quadrant angles is negative. That means the result is a second-quadrant angle, so the cosine will also be negative.
cos(A+B) = -1/√(tan(A+B)² +1)
cos(A+B) = -1/√((1520/1239)² +1)
cos(A+B) = -1239/1961
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Additional comment
Some calculators maintain enough internal accuracy that you can obtain the answer directly from ...
cos(arctan(45/28) +arccos(12/35))
The one shown in the attachment is not able to provide the ratio of integers equal to the floating point value it computes for this.