If `theta_1,theta_2,theta_3, ,theta_n` are in AP, whose common difference is `d` , show that `sectheta_1s e ctheta_2+s e ctheta_2sectheta_3++sectheta_(n-1)sectheta_n=(tantheta_n-tantheta_1)/(sind)`

Explanation

Since `theta _(1), theta_(2) , theta_(3),***** Theta_(n)` are in AP
`theta_(2) - theta_(1)= theta_(3)-theta_(2) =***= theta_(n)-theta_(n-1)=d`
Now , we have to prove
`secq_(1) secq_(2)+secq_(2)secq_(3)+...+secq_(n-1)sectheta_(n)=(tan theta_(n)-tan theta_(1))/(sind)`
Or it can be writen as
`sin d[ sec theta _(1)sec theta_(2) + sectheta_(2)sectheta_(3)+...+ sec theta _(n-1) sec theta_(n)] tan theta_(n)- tan theta_(1)`
Now taking only first term of LHS
`sin d sectheta_(1)sectheta_(2)=(sind)/(costheta_(1)costheta_(2))=sin(theta_(2)-theta_(1))/(cos theta_(1)costheta_(2)) ["from Eq.(i) "]`
` =(sin theta_(2)cos theta_(1)-costheta_(2) sin theta_(1))/( cos theta_(1)cos theta_(2))`
`[:' sin (A-B )=sin A .cos B - cosA . sin B ]`
` =( sin theta_(2)costheta_(1))/(costhet_(1)costheta_(2))-(cos theta_(2)sin theta_(1))/(costheta_(1)costheta(2)) = tan theta_(2)-tan theta_(1)`
Similarly , we can solve other term s which will be tan ` theta _(3)- tan theta_(2), tan theta_(4)-tan theta_(3)....`
`therefore LHs - tan theta_(2)- tantheta_(1) _ tan theta_(3) - tan theta_(2) +...+ tantheta_(n)- tan theta_(n-1)`
`=- tan theta_(1) tan theta_(n)= tan theta _(n)- tan theta_(1)`
=RHS hence proved.