Question

Question #b70e4

Answer 1

`LHS=(sin (n+1)a-sin(n-1)a)/(cos(n+1)a +2cos(n)a+cos(n-1)a)`

`=(2cos(na)sin(a))/(2cos(na)cos(a) +2cos(na))`

`=(2cos(na)sin(a))/(2cos(na)(cos(a) +1)`

`=(2cos(a/2)sin(a/2))/(2cos^2(a/2))`

`=tan(a/2)=RHS`

Proved