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Author: Tinku Tara

4-arctan-1-5-arctan-1-239-

Question Number 126499 by liberty last updated on 21/Dec/20 $$\:\:\:\:\mathrm{4}\:\mathrm{arctan}\:\left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{arctan}\:\left(\frac{\mathrm{1}}{\mathrm{239}}\right)\:=? \\ $$ Answered by benjo_mathlover last updated on 21/Dec/20 $$\:{let}\:{x}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)\:\Rightarrow\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\:\mathrm{tan}\:\mathrm{2}{x}=\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}\:=\:\frac{\mathrm{2}/\mathrm{5}}{\mathrm{1}−\mathrm{1}/\mathrm{25}}\: \\…

Question-60955

Question Number 60955 by Tawa1 last updated on 27/May/19 Commented by alphaprime last updated on 27/May/19 hello sir , I want you to join my community to resolve the question no. 60723 , so please provide me your email Commented by Tawa1 last updated on 27/May/19 $$\mathrm{How}\:?…

Question-192024

Question Number 192024 by Shlock last updated on 05/May/23 Answered by a.lgnaoui last updated on 05/May/23 $$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{shaded}}\:\boldsymbol{\mathrm{Area}}\:\boldsymbol{\mathrm{betwen}} \\ $$$$\left[\boldsymbol{\mathrm{y}}=\mathrm{0},\boldsymbol{\mathrm{y}}=\sqrt{\mathrm{25}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:−\mathrm{3}\:\:;\boldsymbol{\mathrm{y}}=\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:}\right] \\ $$$$\int_{\mathrm{0}} ^{\mathrm{4}} \left(\sqrt{\mathrm{25}−\boldsymbol{\mathrm{x}}^{\mathrm{2}}…

Question-192023

Question Number 192023 by Shlock last updated on 05/May/23 Answered by a.lgnaoui last updated on 06/May/23 $$\boldsymbol{\mathrm{graphe}}\left(\mathrm{1}\right) \\ $$$$\boldsymbol{\mathrm{y}}\:_{\mathrm{1}} \:\:\:\boldsymbol{\mathrm{represente}}\:\boldsymbol{\mathrm{raphe}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{quart}}\:\boldsymbol{\mathrm{circle}}\:\left(\boldsymbol{\mathrm{radius}}\:\mathrm{2}\right)\:\boldsymbol{\mathrm{centre}}\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<\boldsymbol{\mathrm{x}}<\mathrm{2}\:\:\:\boldsymbol{\mathrm{y}}_{\mathrm{1}} =\sqrt{\mathrm{4}−\boldsymbol{\mathrm{x}}^{\mathrm{2}}…

Question-60948

Question Number 60948 by Kunal12588 last updated on 27/May/19 Commented by Prithwish sen last updated on 27/May/19 $$\mathrm{Let}\:\mathrm{A}\:\mathrm{be}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{of}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{plane}\: \\ $$$$\mathrm{and}, \\ $$$$\mathrm{AB}=\mathrm{2a} \\ $$$$\mathrm{BC}=\mathrm{2b} \\…