SOLUTION: If tan 𝛼 = − 12/35 and cot 𝛽 = 3/4 for a second-quadrant angle 𝛼 and a third-quadrant angle 𝛽, find the following. (a) sin(𝛼 + 𝛽) (c) tan(𝛼 + 𝛽

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Question 1199044: If tan 𝛼 = − 12/35 and cot 𝛽 = 3/4
for a second-quadrant angle 𝛼 and a third-quadrant angle 𝛽, find the following.
(a) sin(𝛼 + 𝛽)
(c) tan(𝛼 + 𝛽)
Found 3 solutions by MathLover1, MathTherapy, math_tutor2020:
Answer by MathLover1(20855)   (Show Source): You can put this solution on YOUR website!
If and
for a second-quadrant angle and a third-quadrant angle , find the following:
(a)
(c)


or

or
for a second-quadrant angle :



and


or
or
a third-quadrant angle (beta):


so, use



(a)
(c)

=

=

=



Answer by MathTherapy(10858)   (Show Source): You can put this solution on YOUR website!

If tan 𝛼 = − 12/35 and cot 𝛽 = 3/4
for a second-quadrant angle 𝛼 and a third-quadrant angle 𝛽, find the following.
(a) sin(𝛼 + 𝛽)
(c) tan(𝛼 + 𝛽)
Pay no attention to what that NUT is trying to "sell" you! It's nothing but sheer RUBBISH! Is she ever going to learn?

You may have noticed the following:
1) Length of the hypotenuse can NEVER be negative.
2) , which is NOT , as that person suggests. And, As a matter
of fact, this calculates to 0, although presented in a RIDICULOUS manner (), instead of , etc.
Why would someone have 0 as a denominator? Is that EVER permissible? Some of the things I see on here are as dumb as things can ever be!

If you want correct answers, then read on! 
Required: sin (𝛼 + 𝛽), and: sin (𝛼 + 𝛽) = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽 
. We can see that sin 𝛼, cos 𝛼, sin 𝛽 and cos 𝛽 are needed.

Given:  in the 2nd Quadrant
This is one of the Pytahorean Triples, of the form: 12-35-37. Therefore, hypotenuse = r = 37.
We therefore get: 

           Given:  in the 3rd Quadrant
This is one of the Pytahorean Triples, of the form: 3-4-5. Therefore, hypotenuse = r = 5.
We therefore get: 


So, sin (𝛼 + 𝛽) = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽 
 
       
                    


        

So, cos (𝛼 + 𝛽) = cos 𝛼 cos 𝛽 - sin 𝛼 sin 𝛽

Answer by math_tutor2020(3838)   (Show Source): You can put this solution on YOUR website!

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
Plug in a = 35, b = 12 and solve for c to get c = 37.
I'll let you do these steps.

Diagram:

From this diagram we can determine the following:



--------------------------------------------------
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:


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:



--------------------------------------------------
Continuation of part (a)

To summarize what each diagram gives us:





Then we can use the trig identity shown below










Answer to part (a) is the fraction 104/185


--------------------------------------------------

I'll let you handle part (c).

There are at least two approaches you could take here.

Method 1 is to compute using the trig identity

Then compute and simplify.

Method 2 involves this trig identity


Other approaches could be possible.

A list of identities can be found here
https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

--------------------------------------------------

Edit:

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

This means both and are incorrect as well.
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