Eli F. answered • 02/03/20
I love coffee, teaching, and math!
Okay so this is a complicated multi step problem so stay focused.
First the inequalities are given to show which part, if any, in the fraction is negative. sec(a) is in quadrant 2 meaning cosine is negative or rather the adjacent side of the triangle is negative. cot(b) is also in quadrant 2 this means the cosine part of the function is negative. This means for triangle a the 28 is the adjacent side and is negative, while in triangle b the 21 is negative .
sec(a) = (-35/28) which is equivalent to cos(a) = (-28/35)
With this draw a triangle with -28 as the adjacent side and 35 as the hypotenuse. Then use the Pythagorean theorem to solve for the final side length
*note you can also recognize this triangle as a multiple of the 345 triangle to solve for the final side as well*
a^2 + b^2 = c^2
where we are solving for b^2
b = sqrt(c^2 - a^2) = sqrt(35^2 - (-28)^2) = 21
Now cot(b) = -21/20 which is equivalent to tan(b) = (-20/21)
Now follow the same steps as the previous part so solve for the hypotenuse of this triangle
c = sqrt(a^2 + b^2) = sqrt((-21)^2 + 20^2) = 29
Now cos(a - b) is equivalent to cos(a)cos(b) + sin(a)sin(b)
Using the triangles you drew to describe the two triangles, a and b, substitute each trig function for the corresponding equivalent fractions
Substitute
cos(a) = (-28/35)
cos(b) = (-21/29)
sin(a) = (21/35)
sin(b) = (20/29)
Into
cos(a)cos(b) + sin(a)sin(b) = (-28/35)(-21/29) + (21/35)(20/29)
Simplify to (588/1015) + (420/1015) = (1008/1015)
therefore cos(a - b) = (1008/1015)
The question does not pose what part b of the question is so I will move onto part c
I am using the same substitution as the previous part a
tan(a) = (-21/28)
tan(b) = (-20/21)
tan(a) - tan(b)) (-21/28) - (-20/21) (17/84)
tan(a - b) = -------------------- = -------------------------- = ------------- = (17/144)
1 + tan(a)tan(b) 1 + (-21/28)(-20/21) 1 + (5/7)
Let me know in the comments if there are any typos, I will try and fix them as soon as possible
