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Answer 1 · 2016-08-15T10:27:48

Given
#cosecA=2/sqrt3#
#=>cosecA=cosec(pi/3)=cosec(pi-pi/3)#
(Takling #2pi>A>0#)

#"So "A =pi/3 or (2pi)/3#

So #tanA=tanpi/3=sqrt3#

or #tanA=tan(2pi)/3=tan(pi-pi/3)#
#=-tanpi/3=-sqrt3#

Alternative

Given
#cosecA=2/sqrt3#

#=>(cosecA)^2=(2/sqrt3)^2#

#=>cosec^2A=4/3#

#=>1+cot^2A=4/3#

#=>cot^2A=4/3-1=1/3#

#=>tan^2A=3#

#=>tanA=+-sqrt3#

Given
# cosecA=+ve-># A lying in 1st or 2nd quadrant.
When A is in 1st quadrant then #tanA =sqrt3#

When A is in 2nd quadrant then #tanA =-sqrt3#