using principal value find tan^-1[cot (pi/3)]

We know that  the principal value branch of tan-1 is -π2,π2So tan-1cot π3, = tan-113, as cot π3=13= tan-1tan π6 as tan π6=13=π6And this value lies within  principal value branch of tan-1 so  tan-1cot π3=π6.

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tan-1 [cot(pi/3)] = tan-1 [1/sqrt(3)]

now principal range of y=tan-1 x is -pi/2

so, tan-1[1/sqrt(3)] = pi/6 ... as -pi/2

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TAN -1[COT PIE/3]

TAN-1[1/ROOT3]

PIE/6 i.e 30 DEGREE

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