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tanA-1-n-1-tanB-n-2n-1-A-B-

Question Number 189998 by 073 last updated on 25/Mar/23 $$\mathrm{tanA}=\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$\mathrm{tanB}=\frac{\mathrm{n}}{\mathrm{2n}+\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{A}+\mathrm{B}=?\: \\ $$ Answered by Frix last updated on 25/Mar/23 $${A}=\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}…

we-are-in-N-3-S-p-2-q-2-r-2-q-p-r-24-and-r-lt-p-q-1-Show-that-p-q-r-is-solution-of-S-if-only-r-lt-12-and-p-q-are-solution-of-the-equation-n-2-24-r-n-24-12-r-0-where-

Question Number 124463 by mathocean1 last updated on 03/Dec/20 $${we}\:{are}\:{in}\:\mathbb{N}^{\mathrm{3}} . \\ $$$$\left({S}\right):\:\begin{cases}{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={r}^{\mathrm{2}} }\\{{q}+{p}+{r}=\mathrm{24}\:{and}\:{r}<{p}+{q}}\end{cases} \\ $$$$\left.\mathrm{1}\right)\:{Show}\:{that}\:\left({p};{q};{r}\right)\:{is}\:{solution}\:{of}\:\left({S}\right) \\ $$$${if}\:{only}\:{r}<\mathrm{12}\:{and}\:{p}\:;\:{q}\:{are}\:{solution}\:{of} \\ $$$${the}\:{equation}:\:{n}^{\mathrm{2}} −\left(\mathrm{24}−{r}\right){n}+\mathrm{24}\left(\mathrm{12}−{r}\right)=\mathrm{0}^{} \\ $$$${where}\:{n}\:{is}\:{unknown}.…

2-2-x-3-cos-x-2-1-2-4-x-2-dx-

Question Number 124446 by ZiYangLee last updated on 03/Dec/20 $$\int_{−\mathrm{2}} ^{\mathrm{2}} \left({x}^{\mathrm{3}} \mathrm{cos}\:\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx}=? \\ $$ Answered by Ar Brandon last updated on 03/Dec/20 $$\mathcal{I}=\int_{−\mathrm{2}}…