Question 1183305

Given: 

{{{sin(a) = 5/8}}} and a is in quadrant 2 

 {{{sin(a)  =opp/hyp=5/8}}} =>{{{opp=5}}} and{{{ hyp=8}}}

{{{adj=sqrt(8^2-5^2)}}}
{{{adj=sqrt(64-25)}}}
{{{adj=sqrt(39)}}}

cosine in quadrant 2 is {{{negative}}}

{{{cos(a) =-adj/hyp =-sqrt(39)/8}}}


and {{{sin (b) = -7/9}}} and b is in quadrant 3.

{{{sin (b) = -7/9}}} =>{{{opp=7}}}, {{{hyp=9}}}

{{{adj=sqrt(9^2-7^2)=sqrt(32)=4sqrt(2)}}}

cos in  quadrant 3 is {{{negative}}}

{{{cos(b)=-adj/hyp}}}

{{{cos(b)=-4sqrt(2)/9}}}


so, we have:

{{{sin(a) = 5/8 }}} and {{{cos(a)=-sqrt(39)/8}}}
{{{sin (b) = -7/9}}} and {{{cos(b)=-4sqrt(2)/9}}}


then

{{{tan(a+b) =sin(a + b)/cos(a + b)}}}


{{{tan(a+b) =(sin(a) cos(b) + cos(a) sin(b))/(cos(a) cos(b) - sin(a) sin(b))}}}


{{{tan(a+b) =((5/8) (-4sqrt(2)/9) + (-sqrt(39)/8) (-7/9 ))/((-sqrt(39)/8) (-4sqrt(2)/9) - (5/8) (-7/9))}}}


{{{tan(a+b) =((7sqrt(78) - 40)/(72sqrt(2)))/( (4sqrt(78) + 35)/72)}}}


{{{tan(a+b) =(7sqrt(78) - 40)/(sqrt(2) (35 + 4sqrt(78)))}}}


{{{tan(a+b) =(1/23) (1792sqrt(2) - 405sqrt(39))}}}