সৃজনশীলঃ ‘N=tan−1⁡(cosectan−1⁡x−tan⁡cot−1⁡x)এবং f(θ)=cos⁡θ…’ দিনাজপুর বোর্ড ২০২১ – বিপরীত ত্রিকোনমিতিক ফাংশন ও ত্রিকোণমিতিক সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে,

$N={\tan }^{-1}(\mathrm{cosec}{\tan }^{-1}x-{\mathrm{tancot}}^{-1}x)$

$={\tan }^{-1}(\mathrm{cosec}\mathrm{cosec}-1\frac{\sqrt{1+x^2}}{x}-{\tan }^{-1}\frac{1}{x})$

$={\tan }^{-1}(\frac{\sqrt{1+x^2}}{x}-\frac{1}{x})$

$={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$

$=\frac{1}{2}\cdot 2{\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$

$=\frac{1}{2}\cdot {\tan }^{-1}\frac{2\cdot \frac{\sqrt{1+x^2}-1}{x}}{1-\frac{{(\sqrt{1+x^2}-1)}^2}{x^2}}$

$=\frac{1}{2}{\tan }^{-1}\frac{2\cdot \frac{\sqrt{1+x^2}-1}{x}}{\frac{x^2-\{(1+x^2)-2\sqrt{1+x^2}+1\}}{x^2}}$

$=\frac{1}{2}{\tan }^{-1}\frac{\frac{2\sqrt{1+x^2}-2}{x}}{\frac{x^2-(1+x^2)+2\sqrt{1+x^2}-1}{x^2}}$

$=\frac{1}{2}{\tan }^{-1}\frac{2\sqrt{1+x^2}-2}{x}\cdot \frac{x^2}{2\sqrt{1+x^2}-2}$

$=\frac{1}{2}{\tan }^{-1}x$

$\therefore N=\frac{1}{2}{\tan }^{-1}x$ (দেখানো হলো)

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

$\therefore f\left(2\theta \right)=\cos 2\theta$

$f\left(3\theta \right)=\cos 3\theta$

এখানে, $f\left(\theta \right)+f\left(2\theta \right)+f\left(3\theta \right)=0$; যখন $-2\pi \le \theta \le 2\pi$

বা, $\cos \theta +\cos 2\theta +\cos 3\theta =0$

বা, $\cos 3\theta +\cos \theta +\cos 2\theta =0$

বা, $2\cos \frac{3\theta +\theta }{2}\cos \frac{3\theta -\theta }{2}+\cos 2\theta =0$

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

হয়, $\cos 2\theta =0$

বা, $2\theta =(2n+1)\frac{\pi }{2}$, যেখানে $n\in Z$

$\therefore \theta =(2n+1)\frac{\pi }{4}$

অথবা, $2\cos \theta +1=0$

বা, $2\cos \theta =-1$

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

$\therefore \theta =2n\pi \pm \frac{2\pi }{3}$, যেখানে $n\in Z$

$n=0$ হলে,$\theta =\frac{\pi }{4},\pm \frac{2\pi }{3}$

$n=1$ হলে, $\theta =\frac{3\pi }{4},\frac{8\pi }{3},\frac{4\pi }{3}$

$n=-1$ হলে, $\theta =-\frac{\pi }{4},-\frac{4\pi }{3},-\frac{8\pi }{3}$

$n=2$ হলে, $\theta =\frac{5\pi }{4},\frac{14\pi }{3},\frac{10\pi }{3}$

$n=-2$ হলে, $\theta =-\frac{3\pi }{4},-\frac{10\pi }{3},-\frac{14\pi }{3}$

$n=3$ হলে, $\theta =\frac{7\pi }{4},\frac{20\pi }{3},\frac{16\pi }{3}$

$n=-3$ হলে, $\theta =-\frac{5\pi }{4},-\frac{16\pi }{3},-\frac{20\pi }{3}$

$n=4$ হলে, $\theta =\frac{9\pi }{4},\frac{26\pi }{3},\frac{22\pi }{3}$

$n=-4$ হলে, $\theta =-\frac{7\pi }{4},-\frac{22\pi }{3},-\frac{26\pi }{3}$

$-2\pi \le \theta \le 2\pi$ ব্যবধিতে

$\theta =-\frac{2\pi }{3},-\frac{4\pi }{3},-\frac{3\pi }{4},-\frac{5\pi }{4},-\frac{7\pi }{4},-\frac{\pi }{4},\frac{\pi }{4},\frac{2\pi }{3},\frac{3\pi }{4},\frac{4\pi }{3},\frac{5\pi }{4},\frac{7\pi }{4}$

নির্ণেয় সমাধান :

$\theta =-\frac{2\pi }{3},-\frac{4\pi }{3},-\frac{3\pi }{4},-\frac{5\pi }{4},-\frac{7\pi }{4},-\frac{\pi }{4},\frac{\pi }{4},\frac{2\pi }{3},\frac{3\pi }{4},\frac{4\pi }{3},\frac{5\pi }{4},\frac{7\pi }{4}.$