Question #fc331

Question

2648 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-20T13:30:19

Given

#sinx+siny=1/4#

#=>2sin((x+y)/2)cos((x-y)/2)=1/4......(1)#

Also given

#cosx+cosy=1/3#

#=>2cos((x+y)/2)cos((x-y)/2)=1/3..........(2)#

Dividing (1) by (2) we get

#tan((x+y)/2)=3/4#

Now

#cot(x+y)=1/(tan(x+y))#

#=(1-tan^2((x+y)/2))/(2tan((x+y)/2))#

#=(1-(3/4)^2)/(2xx3/4)#

#=((16-9)/16)/(3/2)#

#=7/16*2/3=7/24#