If `theta_(1),theta_(2),theta_(2), theta_(3),"…………" Theta_(n)` are in AP whose common difference is d, show that sec `theta_(1) sec theta_(2)+ sec_(3) +" …" + sec theta _(n-1)sec theta_(n)=(tantheta_(n)-tan theta_(1))/(sin d )`

Explanation

`theta_(1),theta_(2),theta_(3),".......,"theta_(n)` are in AP.
So, `theta_(2)-theta_(1)=theta_(3)-theta_(2)="......"=theta_(n)-theta_(n-1)=d`
`:.LHS =sind[sectheta_(1) sectheta_(2) sectheta_(2)sectheta_(3)+"....."+sectheta_(n-1)sectheta_(n)]`
`=sind[(1)/(costheta_(1)costheta_(2))+(1)/(costheta_(2)costheta_(3))+"......."+(1)/(costheta_(n-1)costheta_(n))]`
`=(sind)/(costheta_(1)costheta_(2))+(sind)/(costheta_(2)costheta_(3))+"......."+(sind)/(costheta_(n-1)costheta_(n))`
`=(sin(theta_(2)-theta_(1)))/(costheta_(1)costheta_(2))+(sin(theta_(3)-theta_(2)))/(costheta_(2)costheta_(3))+"......."+(sin(theta_(n)-theta_(n-1)))/(costheta_(n-1)costheta_(n))`
`=(tantheta_(2)-tantheta_(1))+(tantheta_(3)-tantheta_(2))+"......."+(tantheta_(n)-tantheta_(n-1))`
`=tantheta_(n)-tantheta_(1)=RHS`.