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Answer 1 · 2017-12-07T05:53:36

# -4/5.#

We have, #sin(u+v)=sinucosv+cosusinv.............(ast).#

Now, since the values of #sinu and cosv# are known; so, to find the

value of #sin(u+v),# we need the values of #cosu, and, sinv.#

Given that #sinu=3/5 rArr cos^2u=1-sin^2u=1-(3/5)^2=16/25.#

#:. cosu=+-4/5," but, as, "u in Q_(II) rArr cosu=-4/5....(ast_1)#.

Similarly, #sinv=+7/25..................................................(ast_2).#

Accordingly, we have, from #(ast),(ast_1) and (ast_2),#

#sin(u+v)=3/5*(-24/25)+(-4/5)(7/25)=(-72-28)/125,#

# rArr sin(u+v)=-100/125=-4/5.#

Enjoy Maths.!