(2sin120+3cos765-2sin240-3cos45)/(5sin300+3tan225-6cos60)
So numerator = 2sin120+3cos765-2sin240-3cos45
and denominator = 5sin300+3tan225-6cos60
Let us solve numerator first
2sin120+3cos765-2sin240-3cos45
or 2sin(90+30) + 3cos(2*360 +45) - 2sin(180+60) - 3cos45
or 2cos(30) + 3cos(45) + 2sin(60) - 3cos(45) ; using sin(90+x) = cosx, cos(n*360 + x) = cosx and sin(180+x) = -sinx
or 2cos(30) + 2sin(60)
or 2sin(60) + 2sin(60) ; using cos(90-x) = sinx
or 4sin(60) ----------------------------(1)
Now we solve denominator
5sin300+3tan225-6cos60
or 5sin(360-60) + 3tan(180+45) - 6cos(60)
or -5sin60 + 3tan45 - 6*1/2 ; since sin(n*360-x) =-sinx, tan(180+x)=tanx and cos60=1/2
or -5sin60 + 3 -3 ; since tan45=1
or -5sin60 ----------------------------------(2)
Now on dividing (1) by (2) we get
4sin60/(-5sin60)
or -4/5
So, on solving (2sin120+3cos765-2sin240-3cos45)/(5sin300+3tan225-6cos60) we get
-4/5 or -0.8