Question #2dd5b

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Answer 1 · 2017-02-15T18:35:19

Given #tanu=5/12#
Both #u and v # are in III quadrant
So
#tanu->+ve#

#tanv->+ve#

#sinu->-ve#

#sinv->-ve#

#cosu->-ve#

#cosv->-ve#

#sinu=1/cscu=-1/sqrt(1+cot^2u)=-1/sqrt(1+12^2/5^2)=-5/13#

#cosu=sinu/tanu=-(5/13)/(5/12)=-12/13#

Given #sinv=-3/5#

#cosv#
#=-sqrt(1-sin^2v)=-sqrt(1-9/25)=-4/5#

#tanv=sinv/cosv=3/4#

#sin(u+v)#

#=sinucosv+cosusinv#

#=(-5/13)(-4/5)+(-12/13)(-3/5)#

#=20/65+36/65=56/65#

#tan(u+v)=(tanu+tanv)/(1-tanutanv)#

#=(5/12+3/4)/(1-5/12*3/4)=(14/12)/(11/16)=7/6*16/11=56/33#

#cos(u+v)#

#=cosucosv-sinusinv#

#=(-12/13)(-4/5)-(-5/13)(-3/5)#

#=48/65-15/65=33/65#