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Question: \(\tan ^ { - 1 } \left( \frac { 1 } { 11 } \right) + \tan ^ { - 1 } \left( \frac { 2 } { 12 } \right...

tan⁡−1(111)+tan⁡−1(212)=\tan ^ { - 1 } \left( \frac { 1 } { 11 } \right) + \tan ^ { - 1 } \left( \frac { 2 } { 12 } \right) =

A

tan⁡−1(33132)\tan ^ { - 1 } \left( \frac { 33 } { 132 } \right)

B

tan⁡−1(12)\tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)

C

tan⁡−1(13233)\tan ^ { - 1 } \left( \frac { 132 } { 33 } \right)

D

None of these

Answer

None of these

Explanation

Solution

tan⁡−1(111)+tan⁡−1(212)\tan ^ { - 1 } \left( \frac { 1 } { 11 } \right) + \tan ^ { - 1 } \left( \frac { 2 } { 12 } \right)

= tan⁡−1(111+2121−111×212)=tan⁡−1(12+22130)\tan ^ { - 1 } \left( \frac { \frac { 1 } { 11 } + \frac { 2 } { 12 } } { 1 - \frac { 1 } { 11 } \times \frac { 2 } { 12 } } \right) = \tan ^ { - 1 } \left( \frac { 12 + 22 } { 130 } \right)

= tan⁡−1(34130)=tan⁡−1(1765)\tan ^ { - 1 } \left( \frac { 34 } { 130 } \right) = \tan ^ { - 1 } \left( \frac { 17 } { 65 } \right) .