সৃজনশীলঃ ‘দৃশ্যকল্প-১: sin⁡α+sin⁡β=aএবং cos⁡α+cos⁡β=bযেখানে,…’ চট্টগ্রাম কলেজ ২০২৫ – সংযুক্ত কোনের ত্রিকোনমিতিক অনুপাত, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়াআছে,

$\sin \alpha +\sin \beta =a$ ............(i)

$\cos \alpha +\cos \beta =b$ ...........(ii)

(i) নং সমীকরণকে (ii) নং সমীকরণ দ্বারা ভাগ করে পাই,

$\therefore \frac{\sin \alpha +\sin \beta }{\cos \alpha +\cos \beta }=\frac{a}{b}$

বা, $\frac{\cos \alpha +\cos \beta }{\sin \alpha +\sin \beta }=\frac{b}{a}$

বা, $\frac{2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}}{2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}}=\frac{b}{a}$

বা, $\frac{{\cos }^2\frac{\alpha +\beta }{2}}{{\sin }^2\frac{\alpha +\beta }{2}}=\frac{b^2}{a^2}$

বা, $\frac{{\cos }^2{90}^{\circ }}{{\sin }^2{90}^{\circ }}=\frac{h^2}{a^2}[\because \alpha +\beta ={180}^{\circ }]$

বা, $\frac{0}{1}=\frac{b^2}{a^2}$

বা, $\frac{b^2}{a^2}=0$

বা, $\frac{b^2-a^2}{b^2+a^2}=\frac{0-1}{0+1}$ [বিয়োজন-যোজনকরে]

বা, $\frac{b^2-a^2}{b^2+a^2}=-1$

$\therefore \frac{b^2-a^2}{b^2+a^2}+1=0$. (প্রমাণিত)

দৃশ্যকল্প-২ হতে, $\cos P(1+{\cos }^{2}P)=\sqrt{2}\cos Q$

বা, $\cos P+{\cos }^{3}P=\sqrt{2}\cos Q$ ............ (i)

এবং$\sin P(1-{\sin }^{2}P)=\sqrt{2}\sin Q$

বা, $\sin P-{\sin }^{3}P=\sqrt{2}\sin Q$ ............ (ii)

(i) নং ও (ii) নং সমীকরণের উভয়পক্ষকে বর্গ করে যোগ করি,

${(\cos P+{\cos }^{3}P)}^2+{(\sin P-{\sin }^{3}P)}^2=2{\cos }^{2}Q+2{\sin }^{2}Q$

বা, ${\cos }^{2}P+2{\cos }^{4}P+{\cos }^{6}P+{\sin }^{2}P-2{\sin }^{4}P+{\sin }^{6}P=2({\cos }^{2}Q+{\sin }^{2}Q)$

বা, $({\cos }^{2}P+{\sin }^{2}P)+2({\cos }^{4}P-{\sin }^{4}P)+{({\cos }^{2}P)}^3+{({\sin }^{2}P)}^3=2$

বা, $1+2({\cos }^{2}P-{\sin }^{2}P)({\cos }^{2}P+{\sin }^{2}P)+({\cos }^{2}P+{\sin }^{2}P)$

$({\cos }^{4}P-{\cos }^{2}P\cdot {\sin }^{2}P+{\sin }^{4}P)=2$

বা, $2({\cos }^{2}P-{\sin }^{2}P)+1\cdot \{{({\cos }^{2}P)}^2+{({\sin }^{2}P)}^2-{\cos }^{2}P\cdot {\sin }^{2}P\}=1$

বা, $2(1-2{\sin }^{2}P)+{({\cos }^{2}P+{\sin }^{2}P)}^2-2{\cos }^{2}P\cdot {\sin }^{2}P-{\cos }^{2}P\cdot {\sin }^{2}P=1$

বা, $2-4{\sin }^{2}P+1-3{\sin }^{2}P\cdot {\cos }^{2}P=1$

বা, $2-4{\sin }^{2}P-3{\sin }^{2}P(1-{\sin }^{2}P)=0$

বা, $.2-4{\sin }^{2}P-3{\sin }^{2}P+3{\sin }^{4}P=0$

বা, $3{\sin }^{4}P-7{\sin }^{2}P+2=0$

বা, $3{\sin }^{4}P-6{\sin }^{2}P-{\sin }^{2}P+2=0$

বা, $3{\sin }^{2}P({\sin }^{2}P-2)-1({\sin }^{2}P-2)=0$

বা, $({\sin }^{2}P-2)(3{\sin }^{2}P-1)=0$

হয়, ${\sin }^{2}P-2=0$

${\sin }^{2}P=2$;

ইহাগ্রহণযোগ্যনয়।

অথবা, $3{\sin }^{2}P-1=0$

${\sin }^{2}P=\frac{1}{3}$

$\therefore {\cos }^{2}P=1-{\sin }^{2}P=1-\frac{1}{3}=\frac{2}{3}$

বামপক্ষ $=\sin (P-Q)=\sin P\cdot \cos Q-\cos P\cdot \sin Q$

$=\sin \left\{\frac{1}{\sqrt{2}}(\cos P+{\cos }^{3}P)\right\}-\cos P\cdot \left\{\frac{1}{\sqrt{2}}(\sin P-{\sin }^{3}P)\right\}$

$=\frac{1}{\sqrt{2}}(\sin P\cdot \cos P+\sin P\cdot {\cos }^{3}P-\sin P\cdot \cos P+{\sin }^{3}P\cdot \cos P)$

$=\frac{1}{\sqrt{2}}\sin P\cdot \cos P({\cos }^{2}P+{\sin }^{2}P)=\frac{1}{\sqrt{2}}\cdot \sin P\cdot \cos P$

$=\frac{1}{\sqrt{2}}\cdot (\pm \frac{1}{\sqrt{3}})\cdot (\pm \frac{\sqrt{2}}{\sqrt{3}})[\because {\sin }^{2}P=\frac{1}{3},{\cos }^{2}P=\frac{2}{3}]$

$=\pm \frac{1}{3}=$ ডানপক্ষ

$\therefore \sin (P-Q)=\pm \frac{1}{3}$. (দেখানো হলো)