Question Number 120468 by help last updated on 31/Oct/20 Commented by bemath last updated on 01/Nov/20 $$\mathrm{cot}\:\left(\theta+\mathrm{30}°\right)\:\mathrm{cot}\:\left(\theta−\mathrm{30}°\right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{tan}\:\left(\theta+\mathrm{30}°\right)\:\mathrm{tan}\:\left(\theta−\mathrm{30}°\right)}\:= \\ $$$$\frac{\mathrm{1}}{\left(\frac{\mathrm{tan}\:\theta+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{1}−\frac{\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta}{\mathrm{3}}}\:.\:\frac{\mathrm{tan}\:\theta−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{1}+\frac{\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta}{\mathrm{3}}}\right)}\:= \\ $$$$\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{2}} \theta}{\mathrm{tan}\:^{\mathrm{2}} \theta−\frac{\mathrm{1}}{\mathrm{3}}}\:=\:\frac{\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}}…
Question Number 120467 by help last updated on 31/Oct/20 Answered by mathmax by abdo last updated on 31/Oct/20 $$\mathrm{tan}\left(\theta+\frac{\pi}{\mathrm{3}}\right)\:=\frac{\mathrm{tan}\theta\:+\mathrm{tan}\frac{\pi}{\mathrm{3}}}{\mathrm{1}−\mathrm{tan}\theta\:\mathrm{tan}\left(\frac{\pi}{\mathrm{3}}\right)}\:=\frac{\mathrm{tan}\theta\:+\sqrt{\mathrm{3}}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{tan}\theta} \\ $$$$\mathrm{tan}\left(\theta+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:=\mathrm{tan}\left(\theta+\pi−\frac{\pi}{\mathrm{3}}\right)\:=\frac{\mathrm{tan}\theta−\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\theta} \\ $$$$\mathrm{e}\:\Rightarrow\mathrm{tan}\theta\:+\frac{\mathrm{tan}\theta\:+\sqrt{\mathrm{3}}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{tan}\theta}\:+\frac{\mathrm{tan}\theta−\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tant}}\:=\mathrm{3}\:\:\:\mathrm{let}\:\mathrm{tan}\theta\:=\mathrm{x}\:\mathrm{so}\:\mathrm{e}\:\Rightarrow \\ $$$$\mathrm{x}\:+\frac{\mathrm{x}+\sqrt{\mathrm{3}}}{\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{x}}+\frac{\mathrm{x}−\sqrt{\mathrm{3}}}{\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{x}}\:=\mathrm{3}\:\Rightarrow\mathrm{x}\:+\frac{\left(\mathrm{x}+\sqrt{\mathrm{3}}\right)\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{x}\right)+\left(\mathrm{x}−\sqrt{\mathrm{3}}\right)\left(\mathrm{1}−\sqrt{\mathrm{3}}\mathrm{x}\right)}{\mathrm{1}−\mathrm{3x}^{\mathrm{2}}…
Question Number 120461 by Khalmohmmad last updated on 31/Oct/20 $$\int\mathrm{sin}\left(\mathrm{ln}{x}\right){dx} \\ $$ Answered by Dwaipayan Shikari last updated on 31/Oct/20 $$\int{sin}\left({logx}\right){dx} \\ $$$$=\left(\int{e}^{{t}} {sin}\left({t}\right){dt}\:\right)\rightarrow{I}\:\:\:\:\:\:\:\:{logx}={t} \\…
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Question Number 54909 by Hassen_Timol last updated on 14/Feb/19 $${Do}\:{you}\:{know}\:{a}\:{geometrical}\:{proof} \\ $$$${of}\:{the}\:{irrationality}\:{of}\:{number}\:\sqrt{\mathrm{2}}\:? \\ $$ Commented by kaivan.ahmadi last updated on 14/Feb/19 https://jeremykun.com/2011/08/14/the-square-root-of-2-is-irrational-geometric-proof/ Commented by kaivan.ahmadi…
Question Number 120445 by zakirullah last updated on 31/Oct/20 Answered by TANMAY PANACEA last updated on 31/Oct/20 $${y}={bx}+{a}\:\:{slope}={b}={tan}\theta=\mathrm{2}.\mathrm{562} \\ $$ Commented by zakirullah last updated…
Question Number 120422 by help last updated on 31/Oct/20 Commented by help last updated on 31/Oct/20 $${Q}\mathrm{5}? \\ $$ Commented by TANMAY PANACEA last updated…
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Question Number 54880 by Otchere Abdullai last updated on 14/Feb/19 $${If}\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:{find}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:{if}\:{a}\:{is} \\ $$$${constant}. \\ $$ Answered by kaivan.ahmadi last updated…
Question Number 120413 by aurpeyz last updated on 31/Oct/20 Commented by JDamian last updated on 31/Oct/20 I cannot understand how the solution is unique when the first force is not clearly defined Answered by TANMAY PANACEA last updated on 31/Oct/20…