Given #sin(-x)=-1/3, tanx=-sqrt2/4# to find the remaining trigonometric function?

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Given #sin(-x)=-1/3, tanx=-sqrt2/4# to find the remaining trigonometric function?

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Answer 1 · 2017-02-04T05:39:20

#sin (-x) = - 1/3 = - sin x# --> #sin x = 1/3#
#cos x = tan x/(sin x) = (-sqrt2/4)(3/1) = - (3sqrt2)/4#
#cot x = - 4/sqrt2 = - (4sqrt2)/2 = - 2sqrt2#
#sec x = 1/(cos) = - 4/(3sqrt2) = - (4sqrt2)/6 = - (2sqrt2)/3#
#csc x = 1/(sin) = 3#

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Given #sin(-x)=-1/3, tanx=-sqrt2/4# to find the remaining trigonometric function?

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