Question Number 193385 by AnshKumar last updated on 12/Jun/23
Answered by som(math1967) last updated on 12/Jun/23
$${L}.{H}.{S} \\ $$$$=\frac{{sec}\mathrm{8}{A}−\mathrm{1}}{{sec}\mathrm{4}{A}−\mathrm{1}} \\ $$$$=\frac{\frac{\mathrm{1}}{{cos}\mathrm{8}{A}}−\mathrm{1}}{\frac{\mathrm{1}}{{cos}\mathrm{4}{A}}−\mathrm{1}} \\ $$$$=\frac{{cos}\mathrm{4}{A}\left(\mathrm{1}−{cos}\mathrm{8}{A}\right)}{{cos}\mathrm{8}{A}\left(\mathrm{1}−{cos}\mathrm{4}{A}\right)} \\ $$$$=\frac{{cos}\mathrm{4}{A}×\mathrm{2}{sin}^{\mathrm{2}} \mathrm{4}{A}}{{cos}\mathrm{8}{A}×\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}{A}} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{4}{Acos}\mathrm{4}{Asin}\mathrm{4}{A}}{\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}{Acos}\mathrm{8}{A}} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{2}{Acos}\mathrm{2}{A}×{sin}\mathrm{8}{A}}{\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}{A}×{cos}\mathrm{8}{A}} \\ $$$$={cot}\mathrm{2}{A}×{tan}\mathrm{8}{A} \\ $$$$=\frac{{tan}\mathrm{8}{A}}{{tan}\mathrm{2}{A}}={R}.{H}.{S} \\ $$