সৃজনশীলঃ ‘m=tan⁡θ,n=sec⁡θ এবং q=cos⁡θ1−sin⁡θ…’ চট্টগ্রাম বোর্ড ২০২৫ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $m=\tan \theta ,n=\sec \theta$ এবং $q=\frac{\cos \theta }{1-\sin \theta }$

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

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

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

$=\sec \theta +\tan \theta$

$=\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }[\because \sec \theta =\frac{1}{\cos \theta }$ এবং $\tan \theta =\frac{\sin \theta }{\cos \theta }]$

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

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

$=\frac{\cos \theta (1+\sin \theta )}{(1+\sin \theta )(1-\sin \theta )}=\frac{\cos \theta }{1-\sin \theta }=q=$ ডানপক্ষ

$\therefore \frac{m+n-1}{m-n+1}=q$. (প্রমাণিত)

দেওয়া আছে, $m=\tan \theta$ এবং $n=\sec \theta$

এখন, $m^2+n^2=\frac{5}{3}$ হলে,

${\tan }^2\theta +{\sec }^2\theta =\frac{5}{3}$

বা, ${\tan }^2\theta +1+{\tan }^2\theta =\frac{5}{3}[\because {\sec }^2\theta =1+{\tan }^2\theta ]$

বা, $2{\tan }^2\theta =\frac{5}{3}-1$

বা, $2{\tan }^2\theta =\frac{2}{3}$

বা, ${\tan }^2\theta =\frac{2}{2\times 3}$

বা, ${\tan }^2\theta =\frac{1}{3}$

$\therefore \tan \theta =\pm \frac{1}{\sqrt{3}}$

$\tan \theta =\frac{1}{\sqrt{3}}$ হলে, ১ম চতুর্ভাগে $\tan \theta =\frac{1}{\sqrt{3}}=\tan \frac{\pi }{6}\therefore \theta =\frac{\pi }{6}$

৩য় চতুর্ভাগে, $\tan \theta =\frac{1}{\sqrt{3}}=\tan (\pi +\frac{\pi }{6})=\tan \frac{7\pi }{6}\therefore \theta =\frac{7\pi }{6}$

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

২য় চতুর্ভাগে, $\tan \theta =-\frac{1}{\sqrt{3}}=-\tan \frac{\pi }{6}=\tan (\pi -\frac{\pi }{6})=\tan \frac{5\pi }{6}$

$\therefore \theta =\frac{5\pi }{6}$

৪র্থ চতুর্ভাগে, $\tan \theta =-\frac{1}{\sqrt{3}}=-\tan \frac{\pi }{6}$

বা, $\tan \theta =-\tan (2\pi -\frac{\pi }{6})=\tan \frac{11\pi }{6}$

$\therefore \theta =\frac{11\pi }{6}$

$0<\theta <2\pi$ ব্যবধিতে $\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$

নির্ণেয় মান : $\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$.