Answer:
![-x^{2} +2xy+10y^{2}](https://tex.z-dn.net/?f=-x%5E%7B2%7D%20%2B2xy%2B10y%5E%7B2%7D)
Step-by-step explanation:
![2x^2-3xy+4y^2-(3x^2-5xy-6y^2)\\2x^2-3xy+4y^2-3x^2+5xy+6y^2\\-x^{2} +2xy+10y^{2}](https://tex.z-dn.net/?f=2x%5E2-3xy%2B4y%5E2-%283x%5E2-5xy-6y%5E2%29%5C%5C2x%5E2-3xy%2B4y%5E2-3x%5E2%2B5xy%2B6y%5E2%5C%5C-x%5E%7B2%7D%20%2B2xy%2B10y%5E%7B2%7D)
The sides could be 1,14 or 2,7
1)
LHS = cot(a/2) - tan(a/2)
= (1 - tan^2(a/2))/tan(a/2)
= (2-sec^2(a/2))/tan(a/2)
= 2cot(a/2) - cosec(a/2)sec(a/2)
= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))
= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)
= 2[(1+cos(a) -1)/sin(a)]
=2cot a
= RHS
2.
LHS = cot(b/2) + tan(b/2)
= [1 + tan^2(b/2)]/tan(b/2)
= sec^2(b/2)/tan(b/2)
= 1/sin(b/2)cos(b/2)
using product to sums
= 2/sin(b)
= 2cosec(b)
= RHS
Add them all up.
A triangle has 3 sides, and you said each side is 3. So:
3+3+3 = 9
321 / 100 = 3.21 as you are dividing by 10, and then 10 again