সৃজনশীলঃ ‘A=sin⁡θ,B=cos⁡θ…’ রাজশাহী বোর্ড ২০২৫ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $A=\sin \theta$ এবং $B=\cos \theta$

বামপক্ষ $=\frac{A-B+1}{A+B-1}=\frac{\sin \theta -\cos \theta +1}{\sin \theta +\cos \theta -1}$

$=\frac{\cos \theta (\frac{\sin \theta }{\cos \theta }-1+\frac{1}{\cos \theta })}{\cos \theta (1+\frac{\sin \theta }{\cos \theta }-\frac{1}{\cos \theta })}=\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}$

$=\frac{\tan \theta +\sec \theta -({\sec }^2\theta -{\tan }^2\theta )}{\tan \theta -\sec \theta +1}[\because {\sec }^2\theta -{\tan }^2\theta =1]$

$=\frac{(\tan \theta +\sec \theta )-(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )}{\tan \theta -\sec \theta +1}$

$=\frac{(\tan \theta +\sec \theta )(1-\sec \theta +\tan \theta )}{(\tan \theta -\sec \theta +1)}$

$=\tan \theta +\sec \theta =\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }=\frac{\sin \theta +1}{\cos \theta }$

$=\frac{(\sin \theta +1)(1-\sin \theta )}{\cos \theta (1-\sin \theta )}=\frac{1-{\sin }^2\theta }{\cos \theta (1-\sin \theta )}$

$=\frac{{\cos }^2\theta }{\cos \theta (1-\sin \theta )}[\because 1-{\sin }^2\theta ={\cos }^2\theta ]$

$=\frac{\cos \theta }{1-\sin \theta }=\frac{B}{1-A}=$ডানপক্ষ

$\therefore \frac{A-B+1}{A+B-1}=\frac{B}{1-A}$ (প্রমাণিত)

দেওয়া আছে, $A=\sin \theta$ এবং $B=\cos \theta$ 

এখন, $\frac{A}{B}+\frac{B}{A}=\frac{4}{\sqrt{3}}$ হলে,

$\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=\frac{4}{\sqrt{3}}$

বা, $\tan \theta +\cot \theta =\frac{4}{\sqrt{3}}[\because \frac{\sin \theta }{\cos \theta }=\tan \theta$ এবং $\frac{\cos \theta }{\sin \theta }=\cot \theta ]$

বা, $\tan \theta +\frac{1}{\tan \theta }=\frac{4}{\sqrt{3}}[\because \cot \theta =\frac{1}{\tan \theta }]$

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

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

বা; $\sqrt{3}{\tan }^2\theta -4\tan \theta +\sqrt{3}=0$

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

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

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

হয়, $\tan \theta -\sqrt{3}=0$

বা, $\tan \theta =\sqrt{3}$

অথবা, $\sqrt{3}\tan \theta -1=0$

বা, $\sqrt{3}\tan \theta =1$

বা, $\tan \theta =\frac{1}{\sqrt{3}}$

এখন, $\tan \theta =\sqrt{3}$ হলে, প্রথম চতুর্ভাগে, $\tan \theta =\sqrt{3}=\tan \frac{\pi }{3}$

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

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

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

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

প্রথম চতুর্ভাগে, $\tan \theta =\frac{1}{\sqrt{3}}=\tan \frac{\pi }{6}$

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

তৃতীয় চতুর্ভাগে, $\tan \theta =\frac{1}{\sqrt{3}}=\tan (\pi +\frac{\pi }{6})=\tan \frac{7\pi }{6}$

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

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

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