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Question Number 42519 by gyugfeet last updated on 27/Aug/18

cosec2A+cosec4A+cosec8A=cotA−cot8A(prlve ghis)

$${cosec}\mathrm{2}{A}+{cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}={cotA}−{cot}\mathrm{8}{A}\left({prlve}\:{ghis}\right) \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18

cotA−(cosec2A+cosec4A+cosec8A)  cotA−cose2A−(cosec4A+cosec8A)  ((cosA)/(sinA))−(1/(2sinAcosA))−(cosec4A+cosec8A)  ((2cos^2 A−1)/(2sinAcosA))−(cosec4A+cosec8A)  ((cos2A)/(sin2A))−cosec4A−cosec8A  cot2A−cosec4A−cosec8A  ((cos2A)/(sin2A))−(1/(sin4A))−cosec8A  ((2cos^2 2A−1)/(2sin2Acos2A))−cosec8A    ((cos4A)/(sin4A))−(1/(sin8A))  ((2cos^2 4A−1)/(sin8A))  ((cos8A)/(sin8A))  cot8A  proved

$${cotA}−\left({cosec}\mathrm{2}{A}+{cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}\right) \\ $$$${cotA}−{cose}\mathrm{2}{A}−\left({cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}\right) \\ $$$$\frac{{cosA}}{{sinA}}−\frac{\mathrm{1}}{\mathrm{2}{sinAcosA}}−\left({cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}\right) \\ $$$$\frac{\mathrm{2}{cos}^{\mathrm{2}} {A}−\mathrm{1}}{\mathrm{2}{sinAcosA}}−\left({cosec}\mathrm{4}{A}+{cosec}\mathrm{8}{A}\right) \\ $$$$\frac{{cos}\mathrm{2}{A}}{{sin}\mathrm{2}{A}}−{cosec}\mathrm{4}{A}−{cosec}\mathrm{8}{A} \\ $$$${cot}\mathrm{2}{A}−{cosec}\mathrm{4}{A}−{cosec}\mathrm{8}{A} \\ $$$$\frac{{cos}\mathrm{2}{A}}{{sin}\mathrm{2}{A}}−\frac{\mathrm{1}}{{sin}\mathrm{4}{A}}−{cosec}\mathrm{8}{A} \\ $$$$\frac{\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}{A}−\mathrm{1}}{\mathrm{2}{sin}\mathrm{2}{Acos}\mathrm{2}{A}}−{cosec}\mathrm{8}{A} \\ $$$$ \\ $$$$\frac{{cos}\mathrm{4}{A}}{{sin}\mathrm{4}{A}}−\frac{\mathrm{1}}{{sin}\mathrm{8}{A}} \\ $$$$\frac{\mathrm{2}{cos}^{\mathrm{2}} \mathrm{4}{A}−\mathrm{1}}{{sin}\mathrm{8}{A}} \\ $$$$\frac{{cos}\mathrm{8}{A}}{{sin}\mathrm{8}{A}} \\ $$$${cot}\mathrm{8}{A}\:\:{proved} \\ $$$$ \\ $$

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