সৃজনশীলঃ ‘…’ রাজউক উত্তরা মডেল কলেজ ২০২৪ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $△\mathrm{ABC}$

$\angle A+\angle B+\angle C=\pi$

$\therefore A+B+C=\pi$

$\therefore A+B=\pi -C$

$\therefore \cot (A+B)=\cot (\pi -C)$

$=\frac{\cot A\cdot \cot B-1}{\cot A\cot B}=-\cot C$

$=\cot A\cdot \cot B-1=-\cot A\cdot \cot C-\cot B\cdot \cot C$

$\therefore \cot \cdot A\cot B+\cot B\cot C+\cot A\cdot \cot C=1\dots \dots \left(i\right)$

দেওয়া আছে,

$\cot A+\cot B+\cot C=\sqrt{3}$

$\Rightarrow (\cot A+\cot B+\cot C{)}^2=(\sqrt{3}{)}^2$

$\Rightarrow {\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C+2\cot A\cdot \cot B+2\cot B\cdot \cot C+2\cot C$.

$\cot A=3$

$\Rightarrow {\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C+2(\cot A\cdot \cot B+\cot B\cdot \cot C+\cot C\cdot \cot$

$\Rightarrow {\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C+2-3=0$

$\Rightarrow {\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C-1=0$

$\Rightarrow {\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C-\cot A\cdot \cot B-\cot B\cdot \cot C-\cot C\cdot \cot A$

$\Rightarrow 2{\cot }^{2}A+2{\cot }^{2}B+2{\cot }^{2}C-2\cot A\cdot \cot B-2\cot B\cdot \cot C-$

$\Rightarrow {\cot }^{2}A-2\cot A\cdot \cot B+{\cot }^{2}B+{\cot }^{2}B-2\cot B\cdot \cot C+{\cot }^{2}C$

$\Rightarrow (\cot A-\cot B{)}^2+(\cot B-\cot C{)}^2+(\cot C-\cot A{)}^2=0$

দুই বা ততোধিক রাশির বর্গের যোগফল শূণ্য হলে এদের প্রত্যেকের মান শূণ্য।

$(\cot A-\cot B{)}^2=0$

$\Rightarrow \cot A-\cot B=0$

$\Rightarrow \cot A=\cot B$

$\therefore A=B$…… (ii)

এবং

$(\cot B-\cot C{)}^2=0$

$\Rightarrow \cot B-\cot C=0$

$\Rightarrow \cot B=\cot C$

$\therefore B=C$. …… (iii)

(ii) ও (iii) হতে,

$A=B=C$

$\therefore △\mathrm{ABC}$ সমবাহু ত্রিভূজ। (proved)

দেওয়া আছে,

$b^4+c^4+a^4=2a^2(b^2+c^2)$

$\Rightarrow a^4+b^4+c^4-2a^2{b}^2-2a^2{c}^2=0$

$\Rightarrow a^4+b^4+c^4-2a^2{b}^2+2b^2{c}^2-2a^2{c}^2=2b^2{c}^2$

$\Rightarrow {(-a^2)}^2+{\left(b^2\right)}^2+{\left(c^2\right)}^2+2\cdot (-a^2)\cdot b^2+2\cdot b^2\cdot c^2+2\cdot c^2\cdot (-a^2)=2b^2{c}^2$

$\Rightarrow (-a^2+b^2+c^2)=2b^2{c}^2$

$\Rightarrow {(c^2+b^2-a^2)}^2=2b^2{c}^2$

$\Rightarrow c^2+b^2-a^2=\sqrt{2}bc$

$\Rightarrow \frac{c^2+b^2-a^2}{bc}=\pm 2\cdot \frac{\sqrt{2}}{2}$

$\Rightarrow \frac{c^2+b^2-a^2}{2bc}=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \cos A=\pm \frac{1}{\sqrt{2}}$

(+) ve নিয়ে,

$\cos A=\frac{1}{\sqrt{2}}$

$=\cos A=\cos {45}^{\circ }$

$\therefore A={45}^{\circ }$

(-) ve নিয়ে,

$\cos A=-\frac{1}{\sqrt{2}}$

$\Rightarrow \cos A=\cos ({180}^{\circ }-{45}^{\circ })$

$\therefore A={135}^{\circ }$ (proved)}