We have,

$A + B + C =90$ (degree)

Now,

$A+B=90-C \ldots \ldots .(1)$

Then,

$\tan (A+B)=\tan (90-C)(\text { from }(1))$

or, $\tan (A+B)=\tan (90-C) \ldots \ldots(2)$

$As$

$\tan (x+y)=(\tan x+\tan y) /(1-\tan x \tan y) \ldots \ldots(3)$

and, $\tan (90-x)=\cot x \ldots \ldots(4)$

$\cup \operatorname{sing}(3)$ and (4) in $(2),$ we get

$(\tan A+\tan B) /(1-\tan A \tan B)=\cot C$

or, $(\tan A+\tan B) /(1-\tan A \tan B)=1 / \tan C$

or, $(\tan C)(\tan A+\tan B)=(1-\tan A \tan B)$

or, $(\tan C \tan A)+(\tan B \tan C)=(1-\tan A \tan B)$

or, tan $A$ tan $B +\tan B \tan C +\tan C \tan A =1$

Therefore, we have

$\tan A \tan B +\tan B \tan C +\tan C \tan A =1$