Question

If none of the angles $x,y$ and $(x+y)$ is a multiple of $\pi$, then prove that
$\cot (x\pm y)=\frac{\cot x\cot y\pm 1}{\cot y\pm \cot x}$

Answer 1

As $x,y$ and $x+y$ are not the multiple of $\pi$, so $\sin x,\sin y$ and $\sin (x+y)$ are non zero.
(i) $\cot (x+y)=\frac{\cos (x+y)}{\sin (x+y)}$
$=\frac{\cos x\cos y-\sin x\sin y}{\sin x\cos y+\sin y\cos x}$
Dividing numerator and denominator by $\sin x\sin y$, we get
$\cot (x+y)=\frac{\frac{\cos x\cos y}{\sin x\sin y}-\frac{\sin x\sin y}{\sin x\sin y}}{\frac{\sin x\cos y}{\sin x\sin y}+\frac{\sin cosx}{\sin x\sin y}}$
$=\frac{\cot x\cot y-1}{\cot y+\cot x}$
(ii) We have $\cot (x+y)=\frac{\cot x\cot y-1}{\cot y+\cot x}$
Replace $y$ by $-y$, we get
$\cot (x+(-y\left)\right)=\frac{\cot x\cot (-y)-1}{\cot (-y)+\cot x}$
$\therefore \cot (x-y)=\frac{-cotx\cot y-1}{-\cot y+\cot x}$
$=\frac{\cot x\cot y+1}{\cot y-\cot x}$