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Question: \(\tan ^ { - 1 } 1 + \tan ^ { - 1 } 2 + \tan ^ { - 1 } 3 =\)...

tan⁡−11+tan⁡−12+tan⁡−13=\tan ^ { - 1 } 1 + \tan ^ { - 1 } 2 + \tan ^ { - 1 } 3 =

A

π2\frac { \pi } { 2 }

B

π4\frac { \pi } { 4 }

C

0

D

None of these

Answer

None of these

Explanation

Solution

tan⁡−11+tan⁡−12+tan⁡−13\tan ^ { - 1 } 1 + \tan ^ { - 1 } 2 + \tan ^ { - 1 } 3

=tan⁡−11+π+tan⁡−1(5−5)= \tan ^ { - 1 } 1 + \pi + \tan ^ { - 1 } \left( \frac { 5 } { - 5 } \right) =tan⁡−11+π−tan⁡−11=π= \tan ^ { - 1 } 1 + \pi - \tan ^ { - 1 } 1 = \pi.