Question 1199044
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I'll do part (a) to get you started.


𝛼 = alpha
𝛽 = beta


Adj = adjacent side
Opp = opposite side
Hyp = hypotenuse


alpha is in quadrant Q2, which is in the northwest.
Any (x,y) point located here has 
x < 0
y > 0
This will mean the adjacent side will be labeled negative, and opposite side positive


Informally we can think of it like this
x = adjacent
y = opposite


tan = opp/adj
tan(alpha) = -12/35 = 12/(-35)
opp = 12
adj = -35


The hypotenuse is 37 found by using the pythagorean theorem {{{a^2+b^2 = c^2}}}
Plug in a = 35, b = 12 and solve for c to get c = 37.
I'll let you do these steps.


Diagram:
{{{drawing(400,400,-5,5,-5,5,
line(-5-3,0,5+3,0),
line(0,-5-3,0,5+3),
red(line(0,0,-3,4)),
red(line(-3,4,-3,0)),
red(line(-3,0,0,0)),
locate(-1,0.75,alpha),

locate(-4,2,"Opp"),
locate(-2,-0.1,"Adj"),
locate(-1.5,2.4,"Hyp"),

locate(-4,1.5,"12"),
locate(-2,-0.6,"-35"),
locate(-1,1.9,"37"),

locate(5-0.4,5-0.4,"Q1"),
locate(-4.6,5-0.4,"Q2"),
locate(-4.6,-4.6,"Q3"),
locate(5-0.4,-4.6,"Q4"),
locate(0.3,-4,matrix(1,2,"Diagram","not")),
locate(0.3,-4.4,matrix(1,2,"to","scale"))
)}}}
From this diagram we can determine the following:
{{{sin(alpha) = opp/hyp = 12/37}}}
{{{cos(alpha) = adj/hyp = -35/37}}}


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Continuation of part (a)


cot(beta) = 3/4
beta is in Q3, which is in the southwest.


This is when x < 0 and y < 0
So I'll have both the opposite and adjacent sides to be labeled negative.


cot = adj/opp = 3/4 = -3/(-4)
adj = -3
opp = -4
Diagram:
{{{drawing(400,400,-5,5,-5,5,
line(-5-3,0,5+3,0),
line(0,-5-3,0,5+3),
red(line(0,0,-3,-4)),
red(line(-3,-4,-3,0)),
red(line(-3,0,0,0)),
locate(-1,-0.5,beta),

locate(-4,-2,"Opp"),
locate(-2,-0.1,"Adj"),
locate(-1.5,-2.4,"Hyp"),

locate(-4,-1.5,"-4"),
locate(-2,0.6,"-3"),
locate(-1,-1.9,"5"),

locate(5-0.4,5-0.4,"Q1"),
locate(-4.6,5-0.4,"Q2"),
locate(-4.6,-4.6,"Q3"),
locate(5-0.4,-4.6,"Q4"),
locate(0.3,-4,matrix(1,2,"Diagram","not")),
locate(0.3,-4.4,matrix(1,2,"to","scale"))
)}}}


The hypotenuse of 5 units is found using the pythagorean theorem (use a = 3, b = 4 to solve for c). I'll let the student perform these steps.


From that second diagram, we can determine:
{{{sin(beta) = opp/hyp = -4/5}}}
{{{cos(beta) = adj/hyp = -3/5}}}


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Continuation of part (a)


To summarize what each diagram gives us:
{{{sin(alpha) = 12/37}}}
{{{cos(alpha) = -35/37}}}
{{{sin(beta) = -4/5}}}
{{{cos(beta)= -3/5}}}


Then we can use the trig identity shown below 
{{{sin(alpha + beta) = sin(alpha)cos(beta) + cos(alpha)sin(beta)}}}


{{{sin(alpha + beta) = (12/37)*(-3/5) + (-35/37)*(-4/5)}}}


{{{sin(alpha + beta) = -36/185 + 140/185}}}


{{{sin(alpha + beta) = (-36 + 140)/185}}}


{{{sin(alpha + beta) = 104/185}}}


Answer to part (a) is the fraction <font color=red>104/185</font>



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I'll let you handle part (c).


There are at least two approaches you could take here.


Method 1 is to compute {{{cos(alpha+beta)}}} using the trig identity {{{cos(alpha+beta) = cos(alpha)cos(beta)-sin(alpha)sin(beta)}}}


Then compute {{{tan(alpha+beta) = (sin(alpha+beta))/(cos(alpha+beta))}}} and simplify.


Method 2 involves this trig identity
{{{tan(alpha+beta) = (tan(alpha)+tan(beta))/(1-tan(alpha)*tan(beta))}}} 


Other approaches could be possible.


A list of identities can be found here
<a href = "https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf">https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf</a>


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Edit:


This in response to what @MathLover1 posted.
She made an error when going from something like {{{sin(alpha) = 12/37}}} (which is correct) to {{{sin(12/37)}}} which is not correct. 
The two ideas are different. 
Similar errors are made elsewhere.


This means both {{{sin(alpha+beta) = -0.2721971}}} and {{{tan(alpha + beta)=-0.2828782}}} are incorrect as well.
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