If θ1, θ2, θ3, …., θn are in AP whose common difference is d, show that
sec θ1sec θ2 + sec θ2sec θ3 + …. + sec θn–1sec θn = .
Show that: sec θ1sec θ2 + sec θ2sec θ3 + …. + sec θn–1sec θn = .
Given: Given AP is θ1, θ2, θ3, …., θn
a= θ1, a2= θ2 and d= θ2 - θ1= θ3 - θ2= θ4 - θ3=…………= θn - θn - 1
sec θ1sec θ2 + sec θ2sec θ3 + …. + sec θn–1sec θn = + ………… +
Multiply both side by sin d
sin d (sec θ1sec θ2 + sec θ2sec θ3 + …. + sec θn–1sec θn)= + ………… +
[NOTE: sin(x - y)=sinxcosy - cosxsiny, & secθcosθ=1]
By using above formula on R.H.S. , we get
R.H.S. = tanθ2 - tanθ1 + tanθ3 - tanθ2 + tanθ4 - tanθ3 ………….. + tanθn - tanθn - 1
R.H.S. = tanθn - tanθ1 (All the remaining term cancle out)
sin d (sec θ1sec θ2 + sec θ2sec θ3 + …. + sec θn–1sec θn)= tanθn - tanθ1 (Divide sin d on both sides), we get
sec θ1sec θ2 + sec θ2sec θ3 + …. + sec θn–1sec θn= .
HENCE PROVED