Question

Eliminate x and y from the equations -
sin x + sin y = a,
cos x + cos y = b,
tan x + tan y = c.

Answer 1

$a^2+b^2=1+1+2cos(x-y)=4co{s}^2\frac{x-y}{2}$
a = 2 sin$\left(\frac{x+y}{2}\right)cos\left(\frac{x-y}{2}\right)$
$\therefore 4a^2=16si{n}^2\left(\frac{x+y}{2}\right)co{s}^2\left(\frac{x-y}{2}\right)$
$\therefore (a^2+b^2{)}^2-4a^2=16co{s}^2\left(\frac{x-y}{2}\right)$
$[co{s}^2\left(\frac{x+y}{2}\right)-si{n}^2\left(\frac{x+y}{2}\right)]$
$=16co{s}^2\left(\frac{x+y}{2}\right)cosxcosy$
again 2ab = sin 2x + sin 2y + 2sin( x + y )
= 2sin (x + y ) cos (x - y ) + 2 sin (x + y )
= 2sin (x + y ) $\left[2co{s}^2\left(\frac{x-y}{2}\right)\right]$
Dividing (2) and (3)
$\frac{(a^2+b^2{)}^2-4a^2}{2ab}=\frac{4cosxcosy}{sin(x+y)}=\frac{4}{c}$
$\because \frac{sin(x+y)}{cosxcosy}=c$
From the third given relation