ID#6180 HSC Higher Math 2nd CQ (Mymensingh 2025)

ক-এর উত্তর:
প্রদত্ত রাশি: $\sec^{2} (\cot^{-1} 2) - 3\sin^{2} (\cos^{-1} \frac{1}{2})$
$\implies \{1 + \tan^{2} (\cot^{-1} 2)\} - 3\{\sin (\frac{\pi}{3})\}^{2}$
$\implies \{1 + \tan^{2} (\tan^{-1} \frac{1}{2})\} - 3\{\frac{\sqrt{3}}{2}\}^{2}$
$\implies \{1 + (\frac{1}{2})^{2}\} - 3(\frac{3}{4})$
$\implies 1 + \frac{1}{4} - \frac{9}{4}$
$\implies \frac{4 + 1 - 9}{4}$
$\implies \frac{-4}{4}$
$\therefore$ নির্ণেয় মান $-1$।

খ-এর উত্তর:
দেওয়া আছে, $f(x) = \cos x$
$\therefore \frac{\cos x + \cos 3x}{\cos 5x + \cos 7x} = -1$
$\implies \frac{2\cos 2x \cos x}{2\cos 6x \cos x} = -1$
$\implies \frac{\cos 2x}{\cos 6x} = -1$ [ধরি $\cos x \neq 0$]
$\implies \cos 6x = -\cos 2x$
$\implies \cos 6x = \cos (\pi - 2x)$
$\implies 6x = 2n\pi \pm (\pi - 2x)$
ধনাত্মক চিহ্ন নিয়ে, $6x = 2n\pi + \pi - 2x \implies 8x = (2n+1)\pi \implies x = \frac{(2n+1)\pi}{8}$
ঋণাত্মক চিহ্ন নিয়ে, $6x = 2n\pi - \pi + 2x \implies 4x = (2n-1)\pi \implies x = \frac{(2n-1)\pi}{4}$
$\therefore$ সাধারণ সমাধান $x = \frac{(2n+1)\pi}{8}$ অথবা $x = \frac{(2n-1)\pi}{4}$ যেখানে $n \in \mathbb{Z}$।

গ-এর উত্তর:
দেওয়া আছে, $f(\theta) = \cos \theta$ এবং $f(\frac{\pi}{2}-\theta) = \sin \theta$
প্রদত্ত সমীকরণ: $\frac{\cos \theta}{1 + \sin \theta} + \frac{\sin \theta}{\cos \theta} = \frac{2}{\sqrt{3}}$
$\implies \frac{\cos^{2} \theta + \sin \theta (1 + \sin \theta)}{\cos \theta (1 + \sin \theta)} = \frac{2}{\sqrt{3}}$
$\implies \frac{\cos^{2} \theta + \sin \theta + \sin^{2} \theta}{\cos \theta (1 + \sin \theta)} = \frac{2}{\sqrt{3}}$
$\implies \frac{1 + \sin \theta}{\cos \theta (1 + \sin \theta)} = \frac{2}{\sqrt{3}}$ [$\because \sin^{2} \theta + \cos^{2} \theta = 1$]
$\implies \frac{1}{\cos \theta} = \frac{2}{\sqrt{3}}$
$\implies \cos \theta = \frac{\sqrt{3}}{2} = \cos \frac{\pi}{6}$
$\therefore \theta = 2n\pi \pm \frac{\pi}{6}$
$-\pi < \theta \le \pi$ ব্যবধিতে:
$n = 0$ হলে, $\theta = \frac{\pi}{6}, -\frac{\pi}{6}$
$n = 1$ হলে, $\theta = 2\pi \pm \frac{\pi}{6}$ [সীমার বাইরে]
$n = -1$ হলে, $\theta = -2\pi \pm \frac{\pi}{6}$ [সীমার বাইরে]
$\therefore$ নির্ণেয় সমাধান $\theta = \frac{\pi}{6}, -\frac{\pi}{6}$।