সৃজনশীলঃ ‘cosecθ−cot⁡θ=1x; যেখানে θ…’ ঢাকা রেসিডেনসিয়াল মডেল কলেজ ২০২৬ – ত্রিকোণমিতি, উচ্চতর গণিত (এসএসসি)

‘ক’ হতে প্রাপ্ত, $\mathrm{cosec}\theta +\cot \theta =x$

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

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

বা, $\frac{(1+\cos \theta {)}^2}{{\sin }^2\theta }=x^2$ [বর্গ করে]

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

বা, $\frac{(1+\cos \theta )(1+\cos \theta )}{(1+\cos \theta )(1-\cos \theta )}=x^2$

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

বা, $\frac{1+\cos \theta -1+\cos \theta }{1+\cos \theta +1-\cos \theta }=\frac{x^2-1}{x^2+1}$ [বিয়োজন-যোজন করে]

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

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

$\therefore \sec \theta =\frac{x^2+1}{x^2-1}$. (দেখানো হলো)

‘খ’ হতে প্রাপ্ত, $\sec \theta =\frac{x^2+1}{x^2-1}$

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

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

বা, ${\cos }^2\theta ={\left(\frac{x^2-1}{x^2+1}\right)}^2$ [বর্গ করে]

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

বা, $1-\frac{{(x^2-1)}^2}{{(x^2+1)}^2}={\sin }^2\theta$

বা, $\frac{{(x^2+1)}^2-{(x^2-1)}^2}{{(x^2+1)}^2}={\sin }^2\theta$

বা, $\frac{4x^2}{{(x^2+1)}^2}={\sin }^2\theta$

বা, ${\sin }^2\theta =\frac{4x^2}{{(x^2+1)}^2}$

বা, $\sin \theta =\sqrt{\frac{4x^2}{{(x^2+1)}^2}}$

$\therefore \sin \theta =\frac{2x}{x^2+1}$

এখন, $\frac{2x}{1+x^2}+\frac{x^2-1}{x^2+1}=\sqrt{2}$ হলে,

বা, $\sin \theta +\cos \theta =\sqrt{2}$

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

বা, $(\sin \theta {)}^2=(\sqrt{2}-\cos \theta {)}^2$ [বর্গ করে]

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

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

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

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

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

বা, $(\sqrt{2}\cos \theta -1{)}^2=0$

বা, $\sqrt{2}\cos \theta -1=0$

বা, $\sqrt{2}\cos \theta =1$

বা, $\cos \theta =\frac{1}{\sqrt{2}}=\cos \frac{\pi }{4}$

$\therefore \theta =\frac{\pi }{4}$

নির্ণেয় মান : $\theta =\frac{\pi }{4}$.