How do you evaluate #\cot (45^\@ )-\cos (30^\@ )#?

Question

1758 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-22T03:40:19

#cot45-cos30=0.134#

show that

#cotx=cosx/sinx#

#cos45=1/sqrt2#

#sin45=1/sqrt2#

#cos30=sqrt3/2#

#cot45-cos30=cos45/sin45-cos30#

#=(1/sqrt2)/(1/sqrt2)-sqrt3/2=1-sqrt3/2=1-0.866=0.134#