Question #ba874

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Answer 1 · 2017-10-08T15:40:15

#cot a = sqrt (-8/9)#

Given, #sec a/tan a = 1/3#

#rArr (1/cos a)/(sin a/cos a) = 1/3#

#rArr 1/cos a. cos a/sin a = 1/3#

#rArr 1/sin a = 1/3#

#rArr cosec a = 1/3# [ squaring both sides]

#rArr cosec^2 a = 1/9#

#rArr 1 + cot^2 a = 1/9# [# we know, 1 + cot^2 a = cosec^2 a#]

#rArr cot^2 a = 1/9-1#

#rArr cot^2 a = -8/9#

#rArr cot a = sqrt (-8/9)#