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

$sin(A+B)$

$=sinAcosB+cosAsinB$

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

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