সৃজনশীলঃ ‘A=cot⁡θ+cosecθ,B=sec⁡θ−tan⁡θ…’ কুমিল্লা জিলা স্কুল ২০২৬ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

[বি.দ্র.: a এর পরিবর্তে B হবে।]

দেওয়া আছে, $a=\sec \theta -\tan \theta$

$\therefore$ ডানপক্ষ $=\frac{1-a}{1+a}=\frac{1-\sec \theta +\tan \theta }{1+\sec \theta -\tan \theta }$

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

$=\frac{\cos \theta -1+\sin \theta }{\cos \theta +1-\sin \theta }=\frac{\sin \theta (\cot \theta -\mathrm{cosec}\theta +1)}{\sin \theta (\cot \theta +\mathrm{cosec}\theta -1)}$

$=\frac{\cot \theta -\mathrm{cosec}\theta +1}{\cot \theta +\mathrm{cosec}\theta -1}$

$=\frac{(\cot \theta -\mathrm{cosec}\theta )+(\mathrm{cosec}{}^2\theta -{\cot }^2\theta )}{\cot \theta +\mathrm{cosec}\theta -1}$

$=\frac{(\cot \theta -\mathrm{cosec}\theta )+(\mathrm{cosec}\theta +\cot \theta )(\mathrm{cosec}\theta -\cot \theta )}{\cot \theta +\mathrm{cosec}\theta -1}$

$=\frac{(\mathrm{cosec}\theta -\cot \theta )(-1+\mathrm{cosec}\theta +\cot \theta )}{\cot \theta +\mathrm{cosec}\theta -1}$

$=\frac{(\mathrm{cosec}\theta -\cot \theta )(\cot \theta +\mathrm{cosec}\theta -1)}{(\cot \theta +\mathrm{cosec}\theta -1)}$$=\mathrm{cosec}\theta -\cot \theta$

= বামপক্ষ

$\therefore \mathrm{cosec}\theta -\cot \theta =\frac{1-a}{1+a}$ (প্রমাণিত)

[বি.দ্র.: B এর পরিবর্তে A হবে]

দেওয়া আছে, $B=\cot \theta +\mathrm{cosec}\theta$

এখন, $B=\frac{1}{\sqrt{3}}$ হলে, $\cot \theta +\mathrm{cosec}\theta =\frac{1}{\sqrt{3}}$

বা, $\frac{\cos \theta }{\sin \theta }+\frac{1}{\sin \theta }=\frac{1}{\sqrt{3}}[\because \cot \theta =\frac{\cos \theta }{\sin \theta }$ এবং $\mathrm{cosec}\theta =\frac{1}{\sin \theta }]$

বা, $\frac{\cos \theta +1}{\sin \theta }=\frac{1}{\sqrt{3}}$

বা, $\sqrt{3}(\cos \theta +1)=\sin \theta$

বা, $\left\{\sqrt{3}\right(\cos \theta +1){\}}^2={\sin }^2\theta$ [বর্গ করে]

বা, $3({\cos }^2\theta +2\cos \theta +1)={\sin }^2\theta$

বা, $3{\cos }^2\theta +6\cos \theta +3=1-{\cos }^2\theta [\because {\sin }^2\theta =1-{\cos }^2\theta ]$

বা, $3{\cos }^2\theta +6\cos \theta +3-1+{\cos }^2\theta =0$

বা, $4{\cos }^2\theta +6\cos \theta +2=0$

বা, $2(2{\cos }^2\theta +3\cos \theta +1)=0$

বা, $2{\cos }^2\theta +3\cos \theta +1=0$

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, $({\cos }_6\theta +1)(2\cos \theta +1)=0$

হয়, $\cos \theta +1=0$

বা, $\cos \theta =-1$

বা, $\cos \theta =\cos \pi$

$\therefore \theta =\pi$

অথবা, $2\cos \theta +1=0$

বা, $2\cos \theta =-1$

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

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

বা, $\cos \theta =\cos \frac{2\pi }{3},\cos \frac{4\pi }{3}\therefore \theta =\frac{2\pi }{3},\frac{4\pi }{3}$

কিন্তু $\theta =\pi ,\frac{4\pi }{3}$ এর জন্য সমীকরণটি সিদ্ধ হয় না।

নির্ণেয় মান : $\theta =\frac{2\pi }{3}$.