Question

tan^-1\(\sqrt{3}\) – cot^-1(-\(\sqrt{3}\)) is equal to
(A) π
(B) \(\frac { π }{ 2 }\)
(C) 0
(D) 2\(\sqrt{3}\)

Answer 1

(B) \(\frac { π }{ 2 }\)

 

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tan^-1\(\sqrt{3}\) = tan^-1(tan\(\frac { π }{ 3 }\)) = \(\frac { π }{ 3 }\)
and cot^-1\(\sqrt{3}\) = cot^-1(-cot\(\frac { π }{ 6 }\) )
= cot^-1cot ( π – \(\frac { π }{ 6 }\) )
= cot^-1cot\(\frac { 5π }{ 6 }\)
= \(\frac { 5π }{ 6 }\).
since the range of principal value branch of cot^-1 is (0, π).
∴ tan^-1\(\sqrt{3}\) – cot^-1(-\(\sqrt{3}\)) = \(\frac { π }{ 3 }\) – (\(\frac { 5π }{ 6 }\)) = \(\frac { 2π – 5π }{ 6 }\)
= \(\frac { -3π }{ 6 }\) = – \(\frac { π }{ 2 }\)