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Question Number 142307 by qaz last updated on 29/May/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left(\mathrm{tan}\:\mathrm{x}\right)−\mathrm{tan}\:\left(\mathrm{tan}\:\left(\mathrm{tan}\:\mathrm{x}\right)\right)}{\mathrm{tan}\:\mathrm{x}\centerdot\mathrm{tan}\:\left(\mathrm{tan}\:\mathrm{x}\right)\centerdot\mathrm{tan}\:\left(\mathrm{tan}\:\left(\mathrm{tan}\:\mathrm{x}\right)\right)}=? \\ $$ Answered by Dwaipayan Shikari last updated on 29/May/21 $${tanx}\approx{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:\:{or}\:\:{x}\:\:\left({sometimes}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}}…
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Question Number 142301 by 777316 last updated on 29/May/21 Answered by Dwaipayan Shikari last updated on 29/May/21 $$\mathrm{sin}\:\left(\frac{\pi{s}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{s}\right)=\mathrm{sin}\:\left(\frac{\pi{s}}{\mathrm{2}}\right)\frac{\pi}{{sin}\left(\pi{s}\right)\Gamma\left({s}\right)} \\ $$$$\underset{{s}\rightarrow\mathrm{2}{n}} {\mathrm{lim}}=\frac{\pi}{\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{2}}{s}\right)\Gamma\left({s}\right)}=\frac{\pi}{\mathrm{2}\Gamma\left(\mathrm{2}{n}\right)\left(−\mathrm{1}\right)^{{n}} } \\ $$ Terms…
Question Number 142300 by qaz last updated on 29/May/21 $$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{e}^{\mathrm{sin}\:\mathrm{x}} +\mathrm{sin}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}} −\left(\mathrm{e}^{\mathrm{tan}\:\mathrm{x}} +\mathrm{tan}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{x}}} }{\mathrm{x}^{\mathrm{3}} }=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 11228 by uni last updated on 17/Mar/17 Answered by mrW1 last updated on 17/Mar/17 $$\left(_{\mathrm{2}} ^{\mathrm{6}} \right)×\left(_{\mathrm{1}} ^{\mathrm{4}} \right)+\left(_{\mathrm{2}} ^{\mathrm{4}} \right)×\left(_{\mathrm{1}} ^{\mathrm{6}} \right)=\frac{\mathrm{6}×\mathrm{5}}{\mathrm{2}}×\mathrm{4}+\frac{\mathrm{4}×\mathrm{3}}{\mathrm{2}}×\mathrm{6}=\mathrm{96}…
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Question Number 11221 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/Mar/17 Commented by mrW1 last updated on 18/Mar/17 $${center}\:{point}\:{of}\:{the}\:{circle}\:{M}\left({a},{a}\right) \\ $$$${radius}\:{of}\:{the}\:{circle}\:{R}=\left({a}−\mathrm{1}\right)\sqrt{\mathrm{2}} \\ $$$$\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\frac{{R}}{{a}\sqrt{\mathrm{2}}}=\mid\frac{{a}−\mathrm{1}}{{a}}\mid\leqslant\mathrm{1} \\ $$$$\theta=\mathrm{2sin}^{−\mathrm{1}} \left(\mid\frac{{a}−\mathrm{1}}{{a}}\mid\right),\:\mid{a}\mid\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\…