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Question Number 52516 by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$$findthevalue$$$$tan\theta +2tan2\theta +2^{2}tan{2}^2\theta +2^{3}tan{2}^3\theta +..+2^{n-1}tan{2}^{n-1}\theta$$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$$cot\theta -tan\theta$$$$=\frac{1-{tan}^2\theta }{tan\theta }$$$$=2\times \frac{1-{tan}^2\theta }{2tan\theta }=\frac{2}{tan2\theta }=2cot2\theta$$$$tan\theta =cot\theta -2cot2\theta ........................eqn1$$$$2\times tan2\theta =2\times cot2\theta -2\times 2cot{2}^2\theta ......eqn2$$$$2^{2}tan{2}^2\theta =2^{2}cot{2}^2\theta -2^{3}cot{2}^3\theta ...........eqn3$$$$.......$$$$......$$$$2^{n-1}tan{2}^{n-1}\theta =2^{n-1}cot{2}^{n-1}\theta -2^{n}cot{2}^n\theta$$$$nowaddlefthandside=\mathit{S}$$$$whenaddingonlyredcolouredtermsremain$$$$otherstermscancelledeachother..$$$$so\mathit{S}=cot\theta -2^{n}cot{2}^n\theta$$$$$$

Commented by malwaan last updated on 10/Jan/19

great thank you

Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

$$mostwelcome...$$

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