Question

show that
tan^1(a+b/1+ab)+tan^1(b-c/1+bc)+tan^1(c-a/1+ca)=0​

Answer 1

Answer:

Step-by-step explanation:

LHS = $tan^{-1}(\frac{a-b}{1+ab}) + tan^{-1}(\frac{b-c}{1+bc})+ tan^{-1}(\frac{c-a}{1+ca})$

       =$tan^{-1}a -tan^{-1}b+tan^{-1}b-tan^{-1}c+tan^{-1}c-tan^{-1}a$

       =0

       =RHS.

FORMULA:      $tan^{-1}x -tan^{-1}y=tan^{-1}(\frac{x-y}{1+xy})$