The given function is tan −1 3 − cot −1 ( − 3 ).
Assume, tan −1 3 =x, then,
3 =tanx 3 =tan π 3 tan −1 3 = π 3
Here π 3 ∈( − π 2 , π 2 ) because ( − π 2 , π 2 ) is the range of principle value branch of tan −1 .
Assume, cot −1 ( − 3 )=y, then,
− 3 =coty =−cot π 6 =cot( π− π 6 ) cot −1 3 = 5π 6
Here 5π 6 ∈( 0,π ) because ( 0,π ) is the range of principle value branch of cot −1 .
Substitute cot −1 3 = 5π 6 and tan −1 3 = π 3 in the given function,
tan −1 3 − cot −1 ( − 3 )= π 3 − 5π 6 = 2π−5π 6 = −3π 6 =− π 2
Therefore, option (B) is correct.