Here,
#a=5 , b=12 and c=13#
#=>BC=5, CA=12 ,AB=13#
#=>bar(AB) # , is the hypotenuse of #triangleABC#
So, #mangleC=90^circ=>angle A and angle B# are acute angles.
Now, #color(red)(cosB=("side adjacent to B")/("hypotenus")=5/13#
Using half angle formula :
#cos(B/2)=(1+cosB)/2=(1+5/13)/2=9/13=(3/sqrt13)^2#
#=>color(red)(cos(B/2)=+3/sqrt13...to[because mangleB <
90^circ]#
Also ,we know the half angle formula:
#color(blue)(tan^2(theta/2)=(1-costheta)/(1+costheta)#
Let us take, #theta=(B/2)#
#:.color(blue)(tan^2((B/2)/2)=(1-cos(B/2))/(1+cos(B/2))#
#=>tan^2(B/4)=(1-3/sqrt13)/(1+3/sqrt13)=(sqrt13-3)/(sqrt13+3)=(sqrt13-3)/(sqrt13+3)xxcolor(green)((1)#
#=>tan^2(B/4)=(sqrt13-3)/(sqrt13+3)xxcolor(green)((sqrt13-3)/(sqrt13-3)#
#=>tan^2(B/4)=(sqrt13-3)^2/((sqrt13)^2-(3)^2)#
#=>tan^2(B/4)=(sqrt13-3)^2/4#
#=>(tan(B/4))^2=((sqrt13-3)/2)^2...tomangleB <90^circ#
#=>tan(B/4)=(sqrt13-3)/2#
