help 19

If tanA=xsinB/[1-xcosB] & tanB=ysinA/[1-ycosA] then prove that -

sinA/sinB= x/y

tanA  = sinA/cosA   and    tanB = sinB/ cosB   ......therefore, we can write :

sinA /cosA  = [xsinB]/ [1 - xcosB]   (1)

sinB/cosB  = [ysinA] / [ 1 - y cosA]  (2)

Rearranging (1)and (2), we have

[sinA] / [sinB]  = [x cosA]/ [1 - xcosB]   (3)

[sinA]/ [sinB] =  [1 - ycosA] / [y cosB]    (4)

Setting the right sides of  (3) and (4) equal gives us

[x cosA]/ [1 - xcosB]   = [1 - ycosA] / [y cosB]     separate the right side

[xcosA] / [1 - xcosB] = [ 1/[ycosB] - cosA/cosB]   cross-multiply

[xcosA]= [1 - xcosB] [1/[ycosB] - cosA/cosB]

[xcosA] = 1/[ycosB] - [x/y] - cosA/cosB + xcosA   

[x/y]   =  1/[ycosB] - [cosA/cosB]     get a common denominator on the right

[x/y]  =  [ 1 - ycosA] / [ ycosB]

And, from (4),  [1 - ycosA]/ [y cosB]  =  [sinA] / [sinB].......therefore,

x/y  =  sinA / sin B