#(sin 120^0 xx csc 150^0 xx tan 225^0) /( sin 330^0 +cos 210^0)#
Assume that all angles are in degrees. Lets we find the values of the mentioned trigonometric ratios:-
#sin 120=Sin (180-60) = sin 60 = sqrt3/2 #----(sine of an angle is positive in the second quadrant.)
#csc150 = csc (180-30) = csc 30= 2 #---(co-secant of an angle is also positive in second quadrant.)
#tan 225= tan (180 + 45)= tan 45= 1 #---(tangent of an angle is positive in third quadrant. )
#sin 330 =sin(360-30) = -sin 30 = -1/2 #---(sine of an angle is negative in fourth quadrant.)
#cos210= cos (180+30) = -cos30= -sqrt3/2# ---(cosine of an angle is negative in third quadrant.)
#rArr(sin 120^0 xx csc 150^0 xx tan 225^0) /( sin 330^0 +cos 210^0)#
#rArr(sqrt3/2 xx 2 xx 1) /( -1/2 -sqrt3/2)#
#rArr(sqrt3 ) /( -1/2 -sqrt3/2)#
#rArr(sqrt3 ) /( (-1 -sqrt3)/2)#
#rArr(-2sqrt3 ) / (1 +sqrt3)#
#rArr(-2sqrt3 ) / (1 +sqrt3) xx (1-sqrt3)/(1-sqrt3)#
#rArr(-2sqrt3+6)/(1-3)#
#rArr(-2sqrt3+6)/-2#
#rArr2sqrt3-3#
