Can anyone evaluate #tan ((5pi)/6) - sin ((7pi)/4) + cot((4pi)/3)# ?

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Answer 1 · 2016-12-17T13:03:06

#tan ((5pi)/6) - sin ((7pi)/4) + cot((4pi)/3)#

#=tan (pi-pi/6) - sin (2pi-pi/4) + cot(pi+pi/3)#

#=-tan (pi/6) - (-sin (pi/4)) - cot(pi/3)#

#=-1/sqrt3+1/sqrt2 -1/sqrt3#

#=-2/sqrt3+1/sqrt2#

#=1/(sqrt6)(sqrt3-2sqrt2)#