Simple Trigo

First draw $\triangle ABC$ where $\angle C=90^\circ$ , $\angle B=30^\circ$ . Extend line $CB$ to $D$ so that $BD=AB$ .

$\angle D=\angle DAB=15^\circ$

So $\angle DAC=75^\circ$

Let $AC=1$ , then $CD=BD+BC=AB+BC=\sqrt3+2$

Hence $\tan 75^\circ=\sqrt3+2$

Now draw $\triangle EFG$ where $\angle G=45^\circ$ , $\angle F=45^\circ$ . Extend line $GF$ to $H$ so that $FH=FE$ .

So $\angle GEH=67.5^\circ$

Let $EG=FG=1$ , then $HG=\sqrt2+1$

Hence $\tan 67.5^\circ=\sqrt2+1$

So, $\tan 75^\circ+\tan 67.5^\circ-\sqrt2-\sqrt3=3$

You can also do it using the double angles formulas for sine and cosine to figure out the tangents of these two angles but you first have to convert 75 and 67. 5 in radians

tan 75 = tan (45 +30) = tan 45 + tan 30 / ( 1 - tan45. tan30)

= 1 + (1/sqrt 3) / (1 - (1/sqrt 3 ))

= 2 + sqrt (3)

tan 135 = - tan 45 = -1

tan 135 = 2 tan 67.5 / 1 - (tan 67.5)^2

then

tan 67.5 = 1 + sqrt (2)

then

tan 75 + tan 67.5 - sqrt (2) - sqrt (3) = 1 + 2 = 3