Question Number 52516 by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19
$${find}\:{the}\:{value} \\ $$$${tan}\theta+\mathrm{2}{tan}\mathrm{2}\theta+\mathrm{2}^{\mathrm{2}} {tan}\mathrm{2}^{\mathrm{2}} \theta+\mathrm{2}^{\mathrm{3}} {tan}\mathrm{2}^{\mathrm{3}} \theta+..+\mathrm{2}^{{n}−\mathrm{1}} {tan}\mathrm{2}^{{n}−\mathrm{1}} \theta \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19
$${cot}\theta−{tan}\theta \\ $$$$=\frac{\mathrm{1}−{tan}^{\mathrm{2}} \theta}{{tan}\theta} \\ $$$$=\mathrm{2}×\frac{\mathrm{1}−{tan}^{\mathrm{2}} \theta}{\mathrm{2}{tan}\theta}=\frac{\mathrm{2}}{{tan}\mathrm{2}\theta}=\mathrm{2}{cot}\mathrm{2}\theta \\ $$$${tan}\theta={cot}\theta−\mathrm{2}{cot}\mathrm{2}\theta........................{eqn}\:\mathrm{1} \\ $$$$\mathrm{2}×{tan}\mathrm{2}\theta=\mathrm{2}×{cot}\mathrm{2}\theta−\mathrm{2}×\mathrm{2}{cot}\mathrm{2}^{\mathrm{2}} \theta......{eqn}\mathrm{2} \\ $$$$\mathrm{2}^{\mathrm{2}} {tan}\mathrm{2}^{\mathrm{2}} \theta=\mathrm{2}^{\mathrm{2}} {cot}\mathrm{2}^{\mathrm{2}} \theta−\mathrm{2}^{\mathrm{3}} {cot}\mathrm{2}^{\mathrm{3}} \theta...........{eqn}\mathrm{3} \\ $$$$....... \\ $$$$...... \\ $$$$\mathrm{2}^{{n}−\mathrm{1}} {tan}\mathrm{2}^{{n}−\mathrm{1}} \theta=\mathrm{2}^{{n}−\mathrm{1}} {cot}\mathrm{2}^{{n}−\mathrm{1}} \theta−\mathrm{2}^{{n}} {cot}\mathrm{2}^{{n}} \theta \\ $$$${now}\:{add}\:{left}\:{hand}\:{side}=\boldsymbol{{S}} \\ $$$${when}\:{adding}\:{only}\:{red}\:{coloured}\:{terms}\:{remain} \\ $$$${others}\:{terms}\:{cancelled}\:{each}\:{other}.. \\ $$$${so}\:\boldsymbol{{S}}={cot}\theta−\mathrm{2}^{{n}} {cot}\mathrm{2}^{{n}} \theta \\ $$$$ \\ $$
Commented by malwaan last updated on 10/Jan/19
great thank you
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19
$${most}\:{welcome}... \\ $$