সৃজনশীলঃ ‘দৃশ্যকল্প-১ : sec⁡A=135,cos⁡…’ ভিকারুননিসা নূন স্কুল এন্ড কলেজ ২০২৬ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $\sec A=\frac{13}{5}$
বা, $A={\sec }^{-1}\frac{13}{5}$
বা, $A={\tan }^{-1}\frac{12}{5}$

$\cos B=\frac{3}{5}$

বা, $B={\cos }^{-1}\frac{3}{5}$

ধরি, ${\cos }^{-1}\frac{3}{5}=\theta$

বা, $\cos \theta =\frac{3}{5}$

বা, $\sin \theta =\frac{4}{5}$

এখন, $\tan \frac{\theta }{2}=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}$

 $=\frac{\sin \theta }{1+\cos \theta }=\frac{\frac{4}{5}}{1+\frac{3}{5}}=\frac{4}{8}=\frac{1}{2}$

$\therefore \frac{\theta }{2}={\tan }^{-1}\frac{1}{2}$

$\therefore \frac{1}{2}B={\tan }^{-1}\frac{1}{2}$

$\tan C=\frac{1}{3}$

বা, $C={\tan }^{-1}\frac{1}{3}$

এবং $\tan D=\frac{142}{31}$

$\therefore D={\tan }^{-1}\frac{142}{31}$

বামপক্ষ $=A-\frac{1}{2}B+2C$

$={\tan }^{-1}\frac{12}{5}-{\tan }^{-1}\frac{1}{2}+2{\tan }^{-1}\frac{1}{3}$

$={\tan }^{-1}\frac{\frac{12}{5}-\frac{1}{2}}{1+\frac{12}{5}\cdot \frac{1}{2}}+{\tan }^{-1}\frac{2\frac{1}{3}}{1-{\left(\frac{1}{3}\right)}^2}$

$={\tan }^{-1}\frac{\frac{24-5}{10}}{\frac{10+12}{10}}+{\tan }^{-1}\frac{\frac{2}{3}}{\frac{9-1}{9}}$

$={\tan }^{-1}\frac{19}{10}\times \frac{10}{22}+{\tan }^{-1}\frac{2}{3}\times \frac{9}{8}$

$={\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{19}{22}$

$={\tan }^{-1}\frac{\frac{3}{4}+\frac{19}{22}}{1-\frac{3}{4}\frac{19}{22}}$

$={\tan }^{-1}\frac{\frac{33+38}{44}}{\frac{88-57}{88}}$

$={\tan }^{-1}\frac{71}{44}\times \frac{88}{31}={\tan }^{-1}\frac{142}{31}$

$=D=$ ডানপক্ষ

$\therefore A-\frac{1}{2}B+2C=D$ .  (প্রমাণিত)

দেওয়া আছে, $f\left(\theta \right)=\sin \theta$

$\therefore f^{\prime }\left(\theta \right)=\cos \theta$
$\therefore f\left(2\theta \right)=\sin 2\theta$
$f(\frac{\pi }{2}-\frac{\theta }{2})=\sin (\frac{\pi }{2}-\frac{\theta }{2})=\cos \frac{\theta }{2}$

প্রদত্ত সমীকরণ,

$2\left[f\right(2\theta )+f(\theta )-{\left\{f(\frac{\pi }{2}-\frac{\theta }{2})\right\}}^2]=f^{\prime }\left(\theta \right)$

বা, $2[\sin 2\theta +\sin \theta -{\{\cos \frac{\theta }{2}\}}^2]=\cos \theta$
বা, $2[2\sin \theta \cos \theta +\sin \theta -{\cos }^2\frac{\theta }{2}]=\cos \theta$
বা, $2[2\sin \theta \cos \theta +\sin \theta -\frac{1}{2}(1+\cos \theta \left)\right]=\cos \theta$
বা, $4\sin \theta \cos \theta +2\sin \theta -1-\cos \theta =\cos \theta$
বা, $4\sin \theta \cos \theta +2\sin \theta -1-2\cos \theta =0$
বা, $2\sin \theta (2\cos \theta +1)-1(2\cos \theta +1)=0$
বা. $(2\cos \theta +1)(2\sin \theta -1)=0$

হয়, $2\cos \theta +1=0$
বা, $\cos \theta =-\frac{1}{2}$
বা, $\theta =2n\pi \pm \frac{2\pi }{3}$,
যেখানে, $n\in Z$

অথবা, $2\sin \theta -1=0$
বা, $\sin \theta =\frac{1}{2}$
বা, $\theta =n\pi +(-1{)}^n\frac{\pi }{6}$,

যেখানে, $n\in Z$

$\therefore \theta =n\pi +(-1{)}^n\frac{\pi }{6},2n\pi \pm \frac{2\pi }{3}$, যেখানে, $n\in Z$
এখন, $n=0$ হলে, $\theta =\frac{\pi }{6},\pm \frac{2\pi }{3}$

$n=-1$ হলে, $\theta =-\frac{7\pi }{6},\frac{-4\pi }{3},\frac{-8\pi }{3}$
$n=1$ হলে, $\theta =\frac{5\pi }{6},\frac{8\pi }{3},\frac{4\pi }{3}$
$n=2$হলে, $\theta =\frac{13\pi }{6},\frac{14\pi }{3},\frac{10\pi }{3}$
$\therefore 0<\theta <\pi$ ব্যবধিতে $\theta =\frac{\pi }{6},\frac{2\pi }{3}$

নির্ণেয় সমাধান : $\theta =\frac{\pi }{6},\frac{2\pi }{3}$.