Answer:
The solution is x=1,y=2,z=3
Step-by-step explanation:
The given system of equations is ;\
2x−3y+4z=8...(1)
3x+4y−5z=−4...(2)
4x−5y+6z=12...(3)
Make x the subject in equation (1)
Put equation (4) into equation (2) and (3)
Multiply through by;
Expand;
Simplify;
Equation (4) in (3)
Put equation (6) into equation (5)
z=3
Put z=3 into equation (6)
y=2(3)-4=2
Put y=2 and z=3 into equation 4
The solution is x=1,y=2,z=3
Answer:
cosx(1-sinx) and ±√(1-cosx)/(1+cosx
Step-by-step explanation:
Given
let x/2 =a then
1-cosx/sinx will be
1-cos2a/sin2a
using cos2x=1-2sin^2x
using sin2x= 2cosxsinx
1-1+2sin^2a/2cosasina
2sina/2cosa
sina/cosa
tana
a=x/2
tana=tanx/2
if x/2=a thensinx/(1+cosx) will be
sin2a/1+cos2a
using sin2x=2cosxsinx and cos2x=-1+2cos^2x
2cosasina/1-1+2cos^2a
2sina/2cosa
sina/cosa
tana
as a=x/2
tana=tanx/2
1-cosx/sinx , sinx/(1+cosx) are identities of tan(x/2)
Both cosx(1-sinx) and ±√(1-cosx)/(1+cosx) are not identities of tan(x/2)!