সৃজনশীলঃ ‘M=sin⁡θ,N=tan⁡θ এবং R=sec⁡θ.…’ বরিশাল বোর্ড ২০২৫ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $M=\sin \theta ,N=\tan \theta$ এবং $R=\sec \theta$

বামপক্ষ $=\frac{1-N-R}{R-N-1}=\frac{1-\tan \theta -\sec \theta }{\sec \theta -\tan \theta -1}$

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

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

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

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

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

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

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

$=\sqrt{\frac{(1+\sin \theta )(1+\sin \theta )}{(1+\sin \theta )(1-\sin \theta )}}$

$=\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}=\sqrt{\frac{1+M}{1-M}}[\because M=\sin \theta ]$ = ডানপক্ষ

$\therefore \frac{1-N-R}{\text{ R- }N-1}=\sqrt{\frac{1+M}{1-M}}$ (প্রমাণিত)

দেওয়া আছে, $M=\sin \theta$ এবং $R=\sec \theta$

এখন, $M^{2}R-\frac{1}{R}=1$ হলে, ${\sin }^2\theta \mathrm{seg}\theta -\frac{1}{\sec \theta }=1$

বা, $\frac{{\sin }^2\theta }{\cos \theta }-\cos \theta =1$

বা, $\frac{{\sin }^2\theta -{\cos }^2\theta }{\cos \theta }=1$

বা, ${\sin }^2\theta -{\cos }^2\theta =\cos \theta$

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

বা, $1-2{\cos }^2\theta =\cos \theta$

বা, $2{\cos }^2\theta +\cos \theta -1=0$

বা, $2{\cos }^2\theta +2\cos \theta -\cos \theta -1=0$

বা, $2\cos \theta (\cos \theta +1)-1(\cos \theta +1)=0$

বা, $(\cos \theta +1)(2\cos \theta -1)=0$

হয়, $\cos \theta +1=0$

বা, $\cos \theta =-1$

বা, $\cos \theta =\cos \pi$

$\therefore \theta =\pi$

অথবা, $2\cos \theta -1=0$

বা, $\cos \theta =\frac{1}{2}=\cos \frac{\pi }{3}=\cos (2\pi -\frac{\pi }{3})$

$\therefore \theta =\frac{\pi }{3},\frac{5\pi }{3}$

$0\le \theta \le 2\pi$ ব্যবধিতে $\theta =\frac{\pi }{3},\frac{5\pi }{3},\pi$

নির্ণেয় মান : $\theta =\frac{\pi }{3},\frac{5\pi }{3},\pi$.