How do you find the value?

Given:
#(180ltalpha<270)#
indicating that #cosalpha # is negative

#tanalpha=40/9#
opposite side is 40
adjacent side is 9
hypotenuse will be 41,
because
#sqrt(9^2+40^2)=sqrt(81+1600)=sqrt1681=41#
#cosalpha="(adjacent side)/(hypotenuse)"#
#cosalpha=-9/41#
#cos(alpha/2)=+-sqrt((1+cosalpha)/2)=+-sqrt((1-9/41)/2#
#=+-sqrt((41-9)/(41xx2))=+-sqrt(32/2xx1/41)=+-sqrt(16)/sqrt(41)#

#cos(alpha/2)=+-4/sqrt41#

As mentioned, #180^@ltalphalt270^@, #
it follows that
#180/2ltalpha/2<270/2#

#90ltalpha/2<135#, where #cos(alpha/2)# is negative

Hence,
#cos(alpha/2)=-4/sqrt41#

Start with the identity:

#tan^2(alpha) +1 = sec^2(alpha)#

Substitute #sec^2(alpha)= 1/cos^2(alpha)#

#tan^2(alpha)+ 1= 1/cos^2(alpha)#

Multiply both sides by #cos^2(alpha)/(tan^2(alpha) + 1)#:

#cos^2(alpha) = 1/(tan^2(alpha) + 1)#

Use the square root operation on both sides:

#cos(alpha) = +-sqrt(1/(tan^2(alpha) + 1))#

We are told that #180^@ < alpha < 270^@#, therefore we choose the negative value:

#cos(alpha) = -sqrt(1/(tan^2(alpha) + 1))#

Add 1 to both sides:

#1+ cos(alpha) = 1-sqrt(1/(tan^2(alpha) + 1))#

Multiply both sides by #1/2#:

#(1+ cos(alpha))/2 = (1-sqrt(1/(tan^2(alpha) + 1)))/2#

Use the square root operation on both sides:

#+-sqrt((1+ cos(alpha))/2) = +-sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)#

Substitute #+-sqrt((1+ cos(alpha))/2)= cos(alpha/2)#

#cos(alpha/2) = +-sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)#

From #180^@ < alpha < 270^@# we derive #90^@ < alpha/2 < 135^@# and conclude that the cosine function is negative within the specified domain:

#cos(alpha/2) = -sqrt((1-sqrt(1/(tan^2(alpha) + 1)))/2)#

Substitute #tan^2(alpha) = (40/9)^2#:

#cos(alpha/2) = -sqrt((1-sqrt(1/((40/9)^2 + 1)))/2)#

I used WolframAlpha to simplify the above into an exact form:

#cos(alpha/2) = -(4sqrt41)/41#