As #cosu=4/7#, #secu=7/4# and squaring #sec^2u=49/16# or
#1+tan^2u=49/16# and #tan^2u=49/16-1=33/16#
and #tanu=sqrt33/4# (note as #u# is in first quadrant it is positive),
Now #sinv=-9/10#, #cosv=sqrt(1-(9/10)^2)#
= #sqrt(1-81/100)=sqrt(19/100)=-sqrt19/10#
and #tanv=sinv/cosv=-9/10xx-10/sqrt19=9/sqrt19# (note as #u# is in third quadrant #tanv# is positive)
Hence, #tan(u-v)=(tanu-tanv)/(1+tanutanv)#
= #(sqrt33/4-9/sqrt19)/(1+sqrt33/4xx9/sqrt19)#
= #(sqrt(19xx33)-36)/(4sqrt19+9sqrt33)#
= #(25.04-36)/(4xx4.3589+9xx5.7446)#
= #(-10.96)/(17.4356+51.7014)=-10/69.137=-0.1446#