If `theta_(1),theta_(2),theta_(3),”…….”,theta_(n)` are in AP whose common difference is d, then show that `sind{sectheta_(1) sectheta_(2)+ sectheta_(2)sectheta_(3)+”……..”+sectheta_(n-1)sectheta_(n)}=tantheta_(n)-tantheta_(1)`.
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`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`.