Question #75cc2

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Answer 1 · 2017-11-27T02:56:52

#sin(A)=4/5, -(3pi)/2 < A < -pi#

#cosA=-sqrt(1-sin^2A)=-sqrt(1-16/25)=-3/5#

Again

# tan(B)=-sqrt(5), pi/2 < B < pi #

#secB=-sqrt(1+tan^2B)#

#=>secB=-sqrt(1+5)=-sqrt6#

#=>cosB=-1/sqrt6#

#sinB=sqrt(1-cos^2B)=sqrt(1-1/6)=sqrt(5/6)#

Now

#sin(A+B)#

#=sinAcosB+cosAsinB#

#=4/5xx(-1/sqrt6)+(-3/5)xxsqrt(5/6)#

#=-((4+3sqrt5)/(5sqrt6))#