Question Number 127793 by Dwaipayan Shikari last updated on 02/Jan/21 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}}{{n}!}{cos}\left(\frac{\pi{n}}{\mathrm{5}}\right) \\ $$ Commented by Dwaipayan Shikari last updated on 03/Jan/21 $${I}\:{have}\:{found} \\…
Question Number 62241 by vishnurajput8081@gmail.com last updated on 18/Jun/19 $$\boldsymbol{{if}}\:\boldsymbol{{the}}\:\boldsymbol{{point}}\:{A}\:{B}\:{C}\:{with}\:{position}\:{vector}\: \\ $$$$\left(\mathrm{20}\hat {{i}}+\lambda\hat {{j}}\right)\:\left(\mathrm{5}\hat {{i}}−\hat {{j}}\right)\:{and}\left(\mathrm{10}\hat {{i}}−\mathrm{13}\hat {{j}}\right)\:{are} \\ $$$${collinear}\:{then}\:{the}\:{value}\:{of}\:\lambda\:{is}: \\ $$ Answered by tanmay…
Question Number 127771 by Dwaipayan Shikari last updated on 01/Jan/21 $$\frac{\mathrm{1}}{\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\pi^{\mathrm{2}} −\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{2}+\pi^{\mathrm{2}} −\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{3}+\pi^{\mathrm{2}} −\frac{\mathrm{4}\pi^{\mathrm{2}} }{\mathrm{4}+\pi^{\mathrm{2}} −\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{5}+\pi^{\mathrm{2}} ….}}}}}} \\ $$ Commented by…
Question Number 127682 by Dwaipayan Shikari last updated on 31/Dec/20 $$\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{3}−\pi^{\mathrm{2}} +\frac{\mathrm{9}\pi^{\mathrm{2}} }{\mathrm{5}−\mathrm{3}\pi^{\mathrm{2}} +\frac{\mathrm{25}\pi^{\mathrm{2}} }{\mathrm{7}−\mathrm{5}\pi^{\mathrm{2}} +\frac{\mathrm{49}\pi^{\:\mathrm{2}} }{\mathrm{9}−\mathrm{7}\pi^{\mathrm{2}} +\frac{\mathrm{81}\pi^{\mathrm{2}} }{\mathrm{11}−\mathrm{9}\pi^{\mathrm{2}} +\frac{\mathrm{121}\pi^{\mathrm{2}} }{…..}}}}}}} \\ $$…
Question Number 62139 by hhghg last updated on 15/Jun/19 $$\mathrm{6}+\mathrm{5}>\mathrm{3}×\mathrm{5}\:\mathrm{true}\:\mathrm{or}\:\mathrm{false} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127666 by Dwaipayan Shikari last updated on 31/Dec/20 $${Have}\:{a}\:{great}\:{year}\:{all}\:{of}\:{you} \\ $$$${It}\:{is}\:\mathrm{12}.\mathrm{20}\:{am}\:{in}\:{India}\:\left({GMT}+\mathrm{5}.\mathrm{30}\right) \\ $$$$\left(\mathrm{12}.\mathrm{20}\right)^{{T}} =\mathrm{20}.\mathrm{21} \\ $$$$ \\ $$💐🌅 Commented by Ar Brandon last…
Question Number 127644 by Dwaipayan Shikari last updated on 31/Dec/20 $$\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}−\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{3}}{\mathrm{2}}−\frac{\frac{\mathrm{1}}{\mathrm{3}}}{\frac{\mathrm{4}}{\mathrm{3}}−\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\frac{\mathrm{5}}{\mathrm{4}}−\frac{\frac{\mathrm{1}}{\mathrm{5}}}{\frac{\mathrm{6}}{\mathrm{5}}−…}}}}}} \\ $$ Commented by Dwaipayan Shikari last updated on 31/Dec/20 $${a}_{\mathrm{0}} +{a}_{\mathrm{0}} {a}_{\mathrm{1}} +{a}_{\mathrm{0}}…
Question Number 62109 by sandhyavs last updated on 15/Jun/19 $${Given}\:{that} \\ $$$$\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}}\right)\mathrm{tan}\:{x}=\left(\mathrm{1}+\sqrt{\mathrm{1}−{x}}\right). \\ $$$${Then}\:{find}\:\:\:\mathrm{sin}\:\mathrm{4}{x}. \\ $$ Commented by sandhyavs last updated on 15/Jun/19 $${Also}\:{tell}\:{me}\:{the}\:{steps}.{Please}… \\…
Question Number 62088 by Cypher1207 last updated on 15/Jun/19 Commented by Cypher1207 last updated on 15/Jun/19 $$\left({very}\right)^{\mathrm{2}} {easy}…. \\ $$ Commented by maxmathsup by imad…
Question Number 62036 by Cypher1207 last updated on 14/Jun/19 $$\mathrm{Correct}\:\mathrm{me}\:\mathrm{if}\:\mathrm{I}\:\mathrm{am}\:\mathrm{wrong}. \\ $$$$\underset{\:\sqrt{−\mathrm{1}}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\mathrm{x}_{\sqrt{−\mathrm{1}}} +\mathrm{y}_{\sqrt{−\mathrm{1}}} \\ $$$$\therefore\left({i}=\sqrt{−\mathrm{1}}\right) \\ $$$$ \\ $$ Commented by Rasheed.Sindhi last…