How do you slove this ?

I would use the fact that:

#tan(x)=sin(x)/cos(x)#

and write it as:

#(1-sin^2(22.5^@)/cos^2(22.5^@))/(sin(22.5^@)/cos(22.5^@))=#

let us rearrange it:

#(cos^2(22.5^@)-sin^2(22.5^@))/cos^cancel(2)(22.5^@)*cancel(cos(22.5^@))/sin(22.5^@)=#

now the tricky part...
Let us multiply and divide by #2#:

#2/2*(cos^2(22.5^@)-sin^2(22.5^@))/(cos(22.5^@)sin(22.5^@))=#

Let us remember some trig identities

that allows us to change top and bottom as:

#2[cos(2*22.5^@)]/sin(2*22.5^@)=2cos(45^@)/sin(45^@)=2*(sqrt(2)/2)/(sqrt(2)/2)=2*1/1=2#

We know that.

#color(blue)(tan^2x=(1-cos2x)/(1+cos2x)#

Take, #x=22.5^circ#

#tan^2 22.5^circ=(1-cos45^circ)/(1+cos45^circ)=(1- 1/sqrt2)/(1+1/sqrt2)=(sqrt2-1)/(sqrt2+1)#

#tan^2 22.5^circ=(sqrt2-1)/(sqrt2+1)xx(sqrt2-1)/(sqrt2-1)= ((sqrt2-1)^2)/(2-1)#

#color(red)(tan^2 22.5^circ=(sqrt2-1)^2=>tan 22.5^circ=sqrt2-1#

Now,

#(1-tg^2 22.5^circ)/(tg22.5^circ)=(1-(sqrt2-1)^2)/(sqrt2-1)#

#(1-tg^2 22.5^circ)/(tg22.5^circ)=(cancel1-2+2sqrt2- cancel1)/(sqrt2-1)=(2sqrt2-2)/(sqrt2-1)#

#(1-tg^2 22.5^circ)/(tg22.5^circ)=(2cancel((sqrt2- 1)))/cancel((sqrt2-1))=2#