সৃজনশীলঃ ‘lsin⁡θ+mcos⁡θ=n এবং c=cot⁡θ.…’ সিলেট বোর্ড ২০২৫ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $l\sin \theta +m\cos \theta =n$

বা, $(l\sin \theta +m\cos \theta {)}^2=n^2$ [বর্গ করে]

বা, $(l\sin \theta {)}^2+2l\sin \theta \cdot m\cos \theta +(m\cos \theta {)}^2=n^2$

বা, $l^2{\sin }^2\theta +2lm\sin \theta \cos \theta +m^2{\cos }^2\theta =n^2$

বা, $l^2(1-{\cos }^2\theta )+2lm\sin \theta \cos \theta +m^2(1-{\sin }^2\theta )=n^2$

$[\because {\sin }^2\theta =1-{\cos }^2\theta$ এবং ${\cos }^2\theta =1-{\sin }^2\theta ]$

বা, $l^2-l^2{\cos }^2\theta +2lm\sin \theta \cos \theta +m^2-m^2{\sin }^2\theta =n^2$

বা, $l^2{\cos }^2\theta -2lm\sin \theta \cos \theta +m^2{\sin }^2\theta =l^2+m^2-n^2$

বা, $(l\cos \theta {)}^2-2l\cos \theta m\sin \theta +(m\sin \theta {)}^2=l^2+m^2-n^2$

বা, $(l\cos \theta -m\sin \theta {)}^2=l^2+m^2-n^2$

$\therefore l\cos \theta -m\sin \theta =\pm \sqrt{l^2+m^2-n^2}$. (প্রমাণিত)

দেওয়া আছে, $c=\cot \theta$

এখানে, $c+\frac{1}{c}=\frac{4}{\sqrt{3}}$

বা, $\cot \theta +\frac{1}{\cot \theta }=\frac{4}{\sqrt{3}}$

বা, $\frac{{\cot }^2\theta +1}{\cot \theta }=\frac{4}{\sqrt{3}}$

বা, $\sqrt{3}{\cot }^2\theta +\sqrt{3}=4\cot \theta$

বা, $\sqrt{3}{\cot }^2\theta -4\cot \theta +\sqrt{3}=0$

বা, $\sqrt{3}{\cot }^2\theta -3\cot \theta -\cot \theta +\sqrt{3}=0$

বা, $\sqrt{3}\cot \theta (\cot \theta -\sqrt{3})-1(\cot \theta -\sqrt{3})=0$

বা, $(\cot \theta -\sqrt{3})(\sqrt{3}\cot \theta -1)=0$

হয়, $\cot \theta -\sqrt{3}=0$

বা, $\cot \theta =\sqrt{3}$

অথবা, $\sqrt{3}\cot \theta -1=0$

বা, $\sqrt{3}\cot \theta =1$

বা, $\cot \theta =\frac{1}{\sqrt{3}}$

এখানে, $\cot \theta =\sqrt{3}$ হলে,

প্রথম চতুর্ভাগে, $\cot \theta =\sqrt{3}=\cot \frac{\pi }{6}$

$\therefore \theta =\frac{\pi }{6}$

তৃতীয় চতুর্ভাগে, $\cot \theta =\sqrt{3}=\cot (\pi +\frac{\pi }{6})=\cot \frac{7\pi }{6}$

$\therefore \theta =\frac{7\pi }{6}$

আবার, $\cot \theta =\frac{1}{\sqrt{3}}$ হলে,

প্রথম চতুর্ভাগে, $\cot \theta =\frac{1}{\sqrt{3}}=\cot \frac{\pi }{3}$

$\therefore \theta =\frac{\pi }{3}$

তৃতীয় চতুর্ভাগে, $\cot \theta =\frac{1}{\sqrt{3}}=\cot (\pi +\frac{\pi }{3})=\cot \frac{4\pi }{3}$

$\therefore \theta =\frac{4\pi }{3}$

$\therefore 0<\theta <2\pi$ ব্যবধিতে $\theta =\frac{\pi }{6},\frac{\pi }{3},\frac{7\pi }{6},\frac{4\pi }{3}$

নির্ণেয় সমাধান : $\theta =\frac{\pi }{6},\frac{\pi }{3},\frac{7\pi }{6},\frac{4\pi }{3}$.