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)
<pre>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) {{{matrix(1,3, tan (alpha + beta), "=", cos (alpha) cos (beta) - sin (alpha) sin (beta))}}}, which is NOT {{{(12/37) * (- 3/5) - (12/37) * (- 3/5)}}}, as that person suggests. And, As a matter
of fact, this calculates to 0, although presented in a RIDICULOUS manner ({{{cos(12/37)}}}), instead of {{{12/37}}}, 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: <font color = red><font size = 4><b>sin (𝛼 + 𝛽)</font></font></b>, and: <font color = red><font size = 4><b>sin (𝛼 + 𝛽) = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽 
</font></font></b>. We can see that sin 𝛼, cos 𝛼, sin 𝛽 and cos 𝛽 are needed.

Given: {{{matrix(1,7, tan (alpha), "=", 12/(- 35), "=", O/A, "=", y/x)}}} in the 2<sup>nd</sup> Quadrant
This is one of the Pytahorean Triples, of the form: 12-35-37. Therefore, hypotenuse = r = 37.
We therefore get: {{{matrix(2,7, sin (alpha), "=", O/H, "=", y/r, "=", 12/37, cos (alpha), "=", A/H, "=", x/r, "=", (- 35)/37)}}}

           Given: {{{matrix(1,9, cot (beta), "=", 3/4, "=", (- 3)/(- 4), "=", A/O, "=", x/y)}}} in the 3<sup>rd</sup> Quadrant
This is one of the Pytahorean Triples, of the form: 3-4-5. Therefore, hypotenuse = r = 5.
We therefore get: {{{matrix(2,7, sin (beta), "=", O/H, "=", y/r, "=", (- 4)/5, cos (beta), "=", A/H, "=", x/r, "=", (- 3)/5)}}}


So, <font color = red><font size = 4><b>sin (𝛼 + 𝛽) = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽 
</font></font></b> 
       {{{matrix(1,3, highlight(sin (alpha + beta)), "=", sin (alpha) cos (beta) + cos (alpha) sin (beta))}}}
                   {{{matrix(3,2, "=", (12/37)(- 3/5) + (- 35/37)(- 4/5), "=", (- 36/185) + (140/185), "=", highlight(104/185))}}} 


        {{{matrix(1,3, tan (alpha + beta), "=", (sin (alpha + beta))/cos (alpha + beta)))}}}

So, <font color = red><font size = 4><b>cos (𝛼 + 𝛽) = cos 𝛼 cos 𝛽 - sin 𝛼 sin 𝛽 
</font></font></b> 
        {{{matrix(1,3, cos (alpha + beta), "=", cos (alpha) cos (beta) - sin (alpha) sin (beta))}}}
                   {{{matrix(3,2, "=", (- 35/37)(- 3/5) - (12/37)(- 4/5), "=", (105/185) - (- 48/185), "=", highlight(153/185))}}} 

       {{{matrix(1,3, highlight(tan (alpha + beta)), "=", (sin (alpha + beta))/cos (alpha + beta)))}}}
                   {{{matrix(4,2, "=", (104/185)/(153/185), "=", (104/185)(185/153), "=", (104/cross(185))(cross(185)/153), "=", highlight(104/153)
)}}}</pre>