Trig2.14

Prove that $\tan (A-B)+\tan (B-C)+\tan (C-A)=\tan (A-B) \tan (B-C) \tan (C-A)\,$

We know that $(A-B)+(B-C)+(C-A)=0\,$

$(A-B)+(B-C)=-(C-A)\,$

Now $\tan [(A-B)+(B-C)]=\tan [-(C-A)]\,$

$\frac{\tan (A-B)+\tan (B-C)}{1-\tan (A-B) \tan (B-C)}=-\tan (C-A)\,$

Cross multiplying, we get

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

$\tan (A-B)+\tan (B-C)+\tan (C-A)=\tan (A-B) \tan (B-C) \tan (C-A)\,$ which is the required.