Trigonometry

[r267747]

1. Prove that:
i) 1+tanAtanA/2=secA=tanAcotA/2-1
ii)tan(x--y)+tan(y-z)+tan(z-x)=tan(x-y)tan(y-z)tan(z-x)
please tell me how to start these questions
2.if cotAcotB=2,show that (cos(A+B))/cos(A-B)=1/3
mY WORK
(cos(A+B))/cos(A-B)=(cosAcosB-sinAsinB)/(cosAcosB+sinAsinB)
PLEASE TELL ME WHAT TO DO NEXT

 

[BigGlenntheHeavy]

$\displaystyle 2) \ Given: \ [cot(A)][cot(B)] \ = \ 2, \ show \ \frac{cos(A+B)}{cos(A-B)} \ = \ \frac{1}{3}$

$\displaystyle [cot(A)][cot(B)] \ = \ \frac{cos(A)cos(B)}{sin(A)sin(B)} \ = \ 2, \ \implies \ cos(A)cos(B) \ = \ 2sin(A)sin(B)$

$\displaystyle \frac{cos(A+B)}{cos(A-B)} \ = \ \frac{cos(A)cos(B)-sin(A)sin(B)}{cos(A)cos(B)+sin(A)sin(B)} \ = \ \frac{cos(A)cos(B)-sin(A)sin(B)}{2sin(A)sin(B)+sin(A)sin(B)}, \ substitution$

$\displaystyle = \ \frac{cos(A)cos(B)-sin(A)sin(B)}{3sin(A)sin(B)} \ = \ \frac{cos(A)cos(B)}{3sin(A)sin(B)}-\frac{sin(A)sin(B)}{3sin(A)sin(B)} \ = \ \frac{2}{3}-\frac{1}{3} \ = \ \frac{1}{3}$