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Author: Tinku Tara

n-1-tan-1-1-2n-2-except-use-tan-1-1-2n-2-tan-1-1-2n-1-tan-1-1-2n-1-any-other-way-

Question Number 137721 by Ñï= last updated on 05/Apr/21 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }=? \\ $$$${except}\:{use}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}},{any}\:{other}\:{way}? \\ $$ Answered by TANMAY…

Solve-this-equation-by-reducing-it-from-non-homogeneous-equation-to-homogeneous-equation-dy-dx-x-y-3-x-y-5-

Question Number 6651 by Tawakalitu. last updated on 08/Jul/16 $${Solve}\:{this}\:{equation}\:{by}\:{reducing}\:{it}\:{from}\:{non}\:{homogeneous} \\ $$$${equation}\:{to}\:{homogeneous}\:{equation}\: \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{x}\:+\:{y}\:+\:\mathrm{3}}{{x}\:−\:{y}\:−\mathrm{5}} \\ $$ Commented by prakash jain last updated on 09/Jul/16 $$\mathrm{Substitue}\:{x}={u}+\mathrm{1}\:\mathrm{and}\:{y}={v}−\mathrm{4}…

What-mass-of-ice-at-14-will-be-needed-to-cool-200-cm-3-of-an-orange-drink-essentially-water-from-25-C-to-10-C-specific-latent-heat-of-fusion-of-ice-3-36-10-5-jkg-1-specific-heat-capac

Question Number 6648 by Tawakalitu. last updated on 07/Jul/16 $${What}\:{mass}\:{of}\:{ice}\:{at}\:\:−\mathrm{14}\:{will}\:{be}\:{needed}\:{to}\:{cool}\:\mathrm{200}\:{cm}^{\mathrm{3}} \:{of} \\ $$$${an}\:{orange}\:{drink}\:\left({essentially}\:{water}\right)\:{from}\:\mathrm{25}°{C}\:\:{to}\:\:\mathrm{10}°{C} \\ $$$$\left({specific}\:{latent}\:{heat}\:{of}\:{fusion}\:{of}\:{ice}\:=\:\mathrm{3}.\mathrm{36}\:×\:\mathrm{10}^{\mathrm{5}} \:{jkg}^{−\mathrm{1}} \right. \\ $$$${specific}\:{heat}\:{capacity}\:{of}\:{ice}\:=\:\mathrm{2100}\:{jkg}^{−\mathrm{1}} {k}^{−\mathrm{1}} \\ $$$${specific}\:{heat}\:{capacity}\:{of}\:{water}\:=\:\mathrm{4200}\:{jkg}^{−\mathrm{1}} {k}^{−\mathrm{1}} . \\…

Question-137717

Question Number 137717 by mnjuly1970 last updated on 05/Apr/21 Commented by Dwaipayan Shikari last updated on 05/Apr/21 $${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{3}−{a}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−{a}}{\mathrm{2}}\right)\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}−{a}}{\mathrm{2}}\right)=\frac{\mathrm{1}−{a}}{\mathrm{4}}.\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}{a}\right)} \\ $$$$=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−{a}\right){sec}\left(\frac{\pi}{\mathrm{2}}{a}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}}{f}'\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{12}} \\ $$ Terms…