Question #90557

Question

1528 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-22T05:55:17

# tan(pi/8)=(sqrt2-1).#

We have,

#tan^2(theta/2)=sin^2(theta/2)/cos^2(theta/2)xx2/2,#

#={2sin^2(theta/2)}/{2cos^2(theta/2)},#

#=(1-costheta)/(1+costheta),#

#:. tan(theta/2)=pmsqrt{(1-costheta)/(1+costheta)}.#

Letting, #theta=pi/4," so that, "theta/2=pi/8,# we have,

#tan(pi/8)=pmsqrt{(1-cos(pi/4))/(1+cos(pi/4))},#

#=pmsqrt{(1-1/sqrt2)/(1+1/sqrt2)},#

#=pmsqrt[{(sqrt2-1)/(sqrt2+1)}xx{(sqrt2-1)/(sqrt2-1)}],#

#=pmsqrt[{(sqrt2-1)^2)/(2-1)],#

#:. tan(pi/8)=pm(sqrt2-1).#

As #(pi/8)" lies in the "1^(st)# Quadrant, it is #+ve.#

# rArr tan(pi/8)=+(sqrt2-1).#

Enjoy Maths.!