How do you find the exact value of the trigonometric function given that #sin u = -sqrt(5) / 6# and #cos v = sqrt(2) / 4# with both u and v in quadrant IV. find sec(u+v) and Tan (u-v)?

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Answer 1 · 2017-09-02T16:34:08

#sec(u+v)=24/(sqrt62-sqrt60)# and #tan(u-v)=(sqrt434-sqrt10)/(sqrt62+sqrt70)#

As #sinu=-sqrt5/6# and #u# is in #Q4#, #cosu# is positive and it is given by #cosu=sqrt(1-(-sqrt5/6)^2)=sqrt(1-5/36)=sqrt31/6#

Further, as #cosv=sqrt2/4# and #v# is in #Q4#, #sinu# is neegative and it is given by #sinv=-sqrt(1-(sqrt2/4)^2)=-sqrt(1-2/16)=-sqrt14/4#

Hence #cos(u+v)=cosucosv-sinusinv#

= #sqrt31/6xxsqrt2/4-(-sqrt5/6)xx(-sqrt14/4)=(sqrt62-sqrt60)/24#

and #sec(u+v)=24/(sqrt62-sqrt60)#

#cos(u-v)=cosucosv+sinusinv#

= #sqrt31/6xxsqrt2/4+(-sqrt5/6)xx(-sqrt14/4)=(sqrt62+sqrt70)/24#

and #sin(u-v)=sinucosv-cosusinv#

= #(-sqrt5/6)xxsqrt2/4-sqrt31/6xx(-sqrt14/4)=(-sqrt10+sqrt434)/24#

= #(sqrt434-sqrt10)/24#

Hence, #tan(u-v)=sin(u-v)/cos(u-v)=(sqrt434-sqrt10)/(sqrt62+sqrt70)#