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

1 Answer
Aug 10, 2018

(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))

=(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))