সৃজনশীলঃ ‘A=cos⁡θ−sin⁡θ+1cos⁡θ+sin⁡θ−1 এবং B=cot⁡θ+cosecθ…’ ময়মনসিংহ বোর্ড ২০২৫ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

দেওয়া আছে, $A=\frac{\cos \theta -\sin \theta +1}{\cos \theta +\sin \theta -1}$

এবং $B=\cot \theta +\mathrm{cosec}\theta$

বামপক্ষ $=A$

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

এবং$=\frac{\cot \theta -1+\mathrm{cosec}\theta }{\cot \theta +1-\mathrm{cosec}\theta }[\because \frac{\cos \theta }{\sin \theta }=\cot \theta \text{ এবং }\frac{1}{\sin \theta }=\mathrm{cosec}\theta ]$

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

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

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

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

$=\cot \theta +\mathrm{cosec}\theta =B=$ ডানপক্ষ

$\therefore A=B$ (দেখানো হলো)

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