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Author: Tinku Tara

Question-21297

Question Number 21297 by NECx last updated on 19/Sep/17 Answered by sma3l2996 last updated on 20/Sep/17 $$\mathrm{2}{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right)+{sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)=\frac{\pi}{\mathrm{2}} \\ $$$${sin}^{−\mathrm{1}} \left(\mathrm{4}{x}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{2}{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{6}}\right) \\ $$$${sin}\left({sin}^{−\mathrm{1}}…

Question-21296

Question Number 21296 by NECx last updated on 19/Sep/17 Commented by NECx last updated on 19/Sep/17 $$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{this}.\:\mathrm{It}\:\mathrm{was} \\ $$$$\mathrm{solved}\:\mathrm{before}\:\mathrm{but}\:\mathrm{i}\:\mathrm{cant}\:\mathrm{assess}\:\mathrm{the} \\ $$$$\mathrm{answer}\:\mathrm{again}. \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}…

Question-152365

Question Number 152365 by mathdanisur last updated on 27/Aug/21 Answered by ghimisi last updated on 28/Aug/21 $${a}={x}+{y};{b}={y}+{z};{c}={x}+{z} \\ $$$${p}={x}+{y}+{z};{q}={ab}+{bc}+{ac};{r}={abc} \\ $$$$\Leftrightarrow….\Leftrightarrow{p}^{\mathrm{3}} +\mathrm{9}{r}\geqslant\mathrm{4}{pq}\Leftrightarrow{schur} \\ $$ Commented…

For-a-particle-performing-uniform-circular-motion-angular-momentum-is-constant-in-magnitude-but-direction-keeps-changing-Am-I-right-or-wrong-

Question Number 21295 by Tinkutara last updated on 19/Sep/17 $$\mathrm{For}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{performing}\:\mathrm{uniform} \\ $$$$\mathrm{circular}\:\mathrm{motion},\:\mathrm{angular}\:\mathrm{momentum}\:\mathrm{is} \\ $$$$\mathrm{constant}\:\mathrm{in}\:\mathrm{magnitude}\:\mathrm{but}\:\mathrm{direction} \\ $$$$\mathrm{keeps}\:\mathrm{changing}. \\ $$$$\mathrm{Am}\:\mathrm{I}\:\mathrm{right}\:\mathrm{or}\:\mathrm{wrong}? \\ $$ Answered by dioph last updated…

Question-152364

Question Number 152364 by mathdanisur last updated on 27/Aug/21 Answered by Kamel last updated on 28/Aug/21 $$ \\ $$$$\Omega\left({a},{b}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{Ln}\left({tan}\left({ax}\right)\right)}{\mathrm{1}−\mathrm{2}{bcos}\left({x}\right)+{b}^{\mathrm{2}} }{dx}\:,\:\mid{b}\mid<\mathrm{1},\:\mathrm{0}<{a}\leqslant\frac{\mathrm{1}}{\mathrm{2}}. \\ $$$${We}\:{have}:\:{Ln}\left({tan}\left({ax}\right)\right)=−\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{+\infty}…