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


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