The value of # (cos (pi/12)-sin (pi/12))(tan (pi/12)+cos( pi/12) )#??

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Answer 1 · 2018-08-10T14:42:28

# (cos (pi/12)-sin (pi/12))(tan (pi/12)+cos( pi/12) )#

#= (cos (pi/12)/cos (pi/12)-sin (pi/12)/cos (pi/12))cos (pi/12)(tan (pi/12)+cos( pi/12) )#

#=(1-tan(pi/12))(sin(pi/12)+cos^2(pi/12))#

#=(1-tan(pi/12))(sin(pi/12)+1/2(1+cos(pi/6))#

Now #tan(pi/12)=tan(pi/3-pi/4)#

#=(tan(pi/3)-tan(pi/4))/(1+tan(pi/3)tan(pi/4))=(sqrt3-1)/(sqrt3+1)#
Again
#sin(pi/12)#

#=sin(pi/3-pi/4)#

#=sin(pi/3)cos(pi/4)-cos(pi/3)sin(pi/4)#

#=(sqrt3-1)/(2sqrt2)#
So
#(1-tan(pi/12))(sin(pi/12)+1/2(1+cos(pi/6))#

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

#=(2/(sqrt3+1))((sqrt3-1)/(2sqrt2)+1/8(4+2sqrt3))#

#=(sqrt3-1)((sqrt3-1)/(2sqrt2)+1/8(sqrt3+1)^2)#

#=((sqrt3-1)^2/(2sqrt2)+1/8(sqrt3-1)(sqrt3+1)^2)#

#=((sqrt3-1)^2/(2sqrt2)+1/4(sqrt3+1))#

#=((4-2sqrt3)/(2sqrt2)+1/4(sqrt3+1))#

#=((2-sqrt3)/(sqrt2)+1/4(sqrt3+1))#

#=((2sqrt2-sqrt6)/2+1/4(sqrt3+1))#