ID#6173 HSC Higher Math 2nd CQ (Rajshahi 2025)

ক-এর উত্তর:
প্রদত্ত রাশি: $\tan^{-1} (\frac{2}{3}) + \sec^{-1} (\frac{\sqrt{13}}{2})$
ধরি, $\sec^{-1} (\frac{\sqrt{13}}{2}) = \theta \implies \sec \theta = \frac{\sqrt{13}}{2}$
$\therefore \tan \theta = \sqrt{\sec^{2} \theta - 1} = \sqrt{(\frac{\sqrt{13}}{2})^{2} - 1} = \sqrt{\frac{13}{4} - 1} = \sqrt{\frac{9}{4}} = \frac{3}{2}$
$\implies \theta = \tan^{-1} (\frac{3}{2})$
এখন, $\tan^{-1} (\frac{2}{3}) + \tan^{-1} (\frac{3}{2}) = \tan^{-1} (\frac{2}{3}) + \cot^{-1} (\frac{2}{3})$
$\because \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$
$\therefore \tan^{-1} (\frac{2}{3}) + \sec^{-1} (\frac{\sqrt{13}}{2}) = \frac{\pi}{2}$ (নির্ণেয় মান)।

খ-এর উত্তর:
দেওয়া আছে, $f(x) = \sin x$
$\therefore f(\frac{\pi}{2}-x) = \sin(\frac{\pi}{2}-x) = \cos x$
প্রদত্ত সমীকরণ: $2\{\cos x\}^{2} + 7\cos x - 4 = 0$
$\implies 2\cos^{2} x + 7\cos x - 4 = 0$
$\implies 2\cos^{2} x + 8\cos x - \cos x - 4 = 0$
$\implies 2\cos x(\cos x + 4) - 1(\cos x + 4) = 0$
$\implies (2\cos x - 1)(\cos x + 4) = 0$
হয়, $2\cos x - 1 = 0 \implies \cos x = \frac{1}{2}$
অথবা, $\cos x + 4 = 0 \implies \cos x = -4$ [যা অসম্ভব, $\because -1 \le \cos x \le 1$]
এখন, $\cos x = \frac{1}{2} = \cos \frac{\pi}{3}$
$\implies x = 2n\pi \pm \frac{\pi}{3}$
$-180^{\circ} \le x \le 180^{\circ}$ ব্যবধিতে:
$n = 0$ হলে, $x = \pm \frac{\pi}{3}$
$n = 1$ হলে, $x = 2\pi \pm \frac{\pi}{3} = \frac{7\pi}{3}, \frac{5\pi}{3}$ [সীমার বাইরে]
$n = -1$ হলে, $x = -2\pi \pm \frac{\pi}{3} = -\frac{5\pi}{3}, -\frac{7\pi}{3}$ [সীমার বাইরে]
$\therefore$ নির্ণেয় সমাধান $x = \frac{\pi}{3}, -\frac{\pi}{3}$।

গ-এর উত্তর:
দেওয়া আছে, $g(x) = \tan x$
প্রদত্ত সমীকরণ: $\{\tan x\}^{2} + 3\tan x - 4 = 0$
$\implies \tan^{2} x + 3\tan x - 4 = 0$
$\implies \tan^{2} x + 4\tan x - \tan x - 4 = 0$
$\implies \tan x(\tan x + 4) - 1(\tan x + 4) = 0$
$\implies (\tan x - 1)(\tan x + 4) = 0$
হয়, $\tan x - 1 = 0 \implies \tan x = 1$
অথবা, $\tan x + 4 = 0 \implies \tan x = -4$
যখন $\tan x = 1 = \tan \frac{\pi}{4}$
$\implies x = n\pi + \frac{\pi}{4}$
যখন $\tan x = -4 = \tan(\tan^{-1}(-4))$
$\implies x = n\pi + \tan^{-1}(-4)$
$\implies x = n\pi - \tan^{-1} 4$
$\therefore$ সাধারণ সমাধান $x = n\pi + \frac{\pi}{4}$ অথবা $x = n\pi - \tan^{-1} 4$ যেখানে $n \in \mathbb{Z}$।