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Question: \(\tan ^ { - 1 } \frac { a - b } { 1 + a b } + \tan ^ { - 1 } \frac { b - c } { 1 + b c } =\)...

tan⁡−1a−b1+ab+tan⁡−1b−c1+bc=\tan ^ { - 1 } \frac { a - b } { 1 + a b } + \tan ^ { - 1 } \frac { b - c } { 1 + b c } =

A

tan⁡−1a−tan⁡−1b\tan ^ { - 1 } a - \tan ^ { - 1 } b

B

tan⁡−1a−tan⁡−1c\tan ^ { - 1 } a - \tan ^ { - 1 } c

C

tan⁡−1b−tan⁡−1c\tan ^ { - 1 } b - \tan ^ { - 1 } c

D

tan⁡−1c−tan⁡−1a\tan ^ { - 1 } c - \tan ^ { - 1 } a

Answer

tan⁡−1a−tan⁡−1c\tan ^ { - 1 } a - \tan ^ { - 1 } c

Explanation

Solution

tan⁡−1(a−b1+ab)+tan⁡−1(b−c1+bc)\tan ^ { - 1 } \left( \frac { a - b } { 1 + a b } \right) + \tan ^ { - 1 } \left( \frac { b - c } { 1 + b c } \right)

=tan⁡−1(a)−tan⁡−1(b)+tan⁡−1(b)−tan⁡−1(c)= \tan ^ { - 1 } ( a ) - \tan ^ { - 1 } ( b ) + \tan ^ { - 1 } ( b ) - \tan ^ { - 1 } ( c )

=tan⁡−1(a)−tan⁡−1(c)= \tan ^ { - 1 } ( a ) - \tan ^ { - 1 } ( c ).