Question

prove the following:
cot-1 {(xy+1) / (x-y)} + cot-1 {(yz+1) / (y-z)} + cot-1 {(zx+1) / (z-x)} = 0

Answer 1

We have to prove:

$\cot^{-1}(\frac{xy+1}{x-y})+ \cot^{-1}(\frac{yz+1}{y-z})+ \cot^{-1}(\frac{zx+1}{z-x})=0$

Let $x = \cot a, y= \cot b , z= \cot c$

$\cot^{-1}(\frac{\cot a \cot b+1}{\cot a-\cot b})+ \cot^{-1}(\frac{\cot b \cot c+1}{\cot b-\cot c})+ \cot^{-1}(\frac{\cot c \cot a+1}{\cot c-\cot a})$

Now, using the trigonometric identity $\cot(a-b) = \frac{ \cot a \cot b+1}{\cot b - \cot a}$.

$\cot^{-1} \cot(b-a)+ \cot^{-1} \cot(c-b)+ \cot^{-1} \cot(a-c)$

b-a+c-b+a-c

= 0

Hence, proved

Answer 2

Answer:

we have to prove that $\cot^{-1}(\frac{xy+1}{x-y})+\cot^{-1}(\frac{yz+1}{y-z})+\cot^{-1}(\frac{zx+1}{z-x})=0$

Left hand side

$\cot^{-1}(\frac{xy+1}{x-y})+\cot^{-1}(\frac{yz+1}{y-z})+\cot^{-1}(\frac{zx+1}{z-x})$

Let $x=\cot a,\; y=\cot b,\; z=\cot c$

Since, by the identity $\cot(a-b)=\frac{\cot a\cot b+1}{\cot b-\cot a}$

$\cot^{-1}\cot(b-a)+\cot^{-1}\cot(c-b)+\cot^{-1}\cot(a-c)$

$b-a+c-b+a-c$

$=0$

Right hand side

hence proved.