How do you evaluate the expression #tan(u-v)# given #cosu=4/7# with #0<u<pi/2# and #sinv=-9/10# with #pi<v<(3pi)/2#?

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Answer 1 · 2016-09-29T09:26:10

#tan(u-v)=-0.1446#

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#