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Question Number 34500 by Neelam last updated on 07/May/18

prove tan^(−1) ((1/2)tan2A)+tan^(−1) (cotA)+tan^(−1) (cot^3 A)=0

$${prove} \\ $$$${t}\mathrm{an}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}{tan}\mathrm{2}{A}\right)+\mathrm{tan}^{−\mathrm{1}} \left({cotA}\right)+\mathrm{tan}^{−\mathrm{1}} \left({cot}^{\mathrm{3}} {A}\right)=\mathrm{0} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/May/18

let tanA=t tan^(−1) ((1/2)×((2t)/(1−t^2 )))+tan^(−1) ((1/t))+tan^(−1) ((1/t^3 )) =tan^(−1) ((((t/(1−t^2 )) +(1/t))/(1−(1/(1−t^(2 _ ) )))))+tan^(−1) ((1/t^3 )) =tan^(−1) (((1/(t−t^3 ))/((1−t^2 −1)/(1−t^2 ))))+tan^(−1) ((1/t^(3 ) )) tan^(−1) {((1−t^2 )/(t(1−t^2 )(−t^2 )))}+tan^(−1) ((1/t^(3 ) )) =tan^(−1) ((1/(−t^3 )))+tan^(−1) ((1/t^3 )} =−tan((1/t^ ))+tan^(−1) ((1/t^3 ) )=0

$${let}\:{tanA}={t} \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} \:}\right)+{tan}^{−\mathrm{1}} \:\left(\frac{\mathrm{1}}{{t}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\frac{{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{t}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}\:_{} \:} \:}}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{{t}−{t}^{\mathrm{3}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }}\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{t}^{\mathrm{3}\:} }\right) \\ $$$${tan}^{−\mathrm{1}} \left\{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left(−{t}^{\mathrm{2}} \right)}\right\}+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{t}^{\mathrm{3}\:} }\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{−{t}^{\mathrm{3}} }\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\right\} \\ $$$$=−{tan}\left(\frac{\mathrm{1}}{{t}^{} }\right)+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\:\right)=\mathrm{0} \\ $$

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