Question

Eliminate $x$ and $y$ from the equations :
$\sin x+\sin y=a$
$\cos x+\cos y=b$ and $\tan x+\tan y=c$

Answer 1

$\sin x+\sin y=a$ ...(1)

$\cos x+\cos y=b$ ...(2)

$\tan x+\tan y=c$ ...(3)

$\Rightarrow a^2+b^2={(\sin x+\sin y)}^2+{(\cos x+\cos y)}^2=2+2\cos (x-y)=4{\cos }^2\left(\frac{x-y}{2}\right)$ ...(4)

Also $a=\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right).\cos \left(\frac{x-y}{2}\right)$

From (4) and (5)

${(a^2+b^2)}^2-4a^2=16{\cos }^4\left(\frac{x-y}{2}\right)-16{\sin }^2\left(\frac{x+y}{2}\right){\cos }^2\left(\frac{x-y}{2}\right)$

$=16{\cos }^2\left(\frac{x-y}{2}\right)\cos x\cos y$

Again $2ab=2(\sin x+\sin y)(\cos x+\cos y)=\sin 2x+\sin 2y+2\sin (x+y)$

$=2\sin (x+y).\cos (x-y)+2\sin (x+y)$

$=2\sin (x+y)(\cos (x-y)+1)=4\sin (x+y){\cos }^2\left(\frac{x-y}{2}\right)$

$\Rightarrow 8ab=16\sin (x+y).{\cos }^2\left(\frac{x-y}{2}\right)$ ...(7)

Dividing (7) by (6), we get

$\frac{8ab}{{(a^2+b^2)}^2-4a^2}=\frac{16\sin (x+y).{\cos }^2\left(\frac{x-y}{2}\right)}{16{\cos }^2\left(\frac{x-y}{2}\right)\cos x\cos y}=\tan x+\tan y=c$