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)`.

If `theta_(1),theta_(2),theta_(3),".......",theta_(n)` are in AP whose

1.	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)`.
Answer» `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`.

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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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