We can use this fundamental trigonometric identity:
#tantheta=sintheta/costheta#
Here's a reference triangle with our #angletheta#:
Since we know #sin(pi/3)# is #sqrt3/2# and #cos(pi/3)# is #1/2#, we can use the previously stated identity to figure out the value of #tan(pi/3)#:
#tan(pi/3)=(quadsin(pi/3)quad)/cos(pi/3)#
#color(white)(tan(pi/3))=(quadsqrt3/2quad)/(1/2)#
#color(white)(tan(pi/3))=sqrt3/2*2/1#
#color(white)(tan(pi/3))=sqrt3/color(red)cancelcolor(black)2*color(red)cancelcolor(black)2/1#
#color(white)(tan(pi/3))=sqrt3/1*1/1#
#color(white)(tan(pi/3))=sqrt3/1*1#
#color(white)(tan(pi/3))=sqrt3/1#
#color(white)(tan(pi/3))=sqrt3#
That's the value of #tan(pi/3)#. Hope this helped!
