ID#6197 HSC Higher Math 2nd CQ (Comilla 2025)

ক-এর উত্তর:
প্রদত্ত রাশি: $\sec^{-1} \frac{\sqrt{34}}{5} + \sin^{-1} \frac{1}{\sqrt{17}}$
$\implies \tan^{-1} \frac{\sqrt{(\sqrt{34})^{2} - 5^{2}}}{5} + \tan^{-1} \frac{1}{\sqrt{(\sqrt{17})^{2} - 1^{2}}}$
$\implies \tan^{-1} \frac{\sqrt{34 - 25}}{5} + \tan^{-1} \frac{1}{\sqrt{17 - 1}}$
$\implies \tan^{-1} \frac{3}{5} + \tan^{-1} \frac{1}{4}$
$\implies \tan^{-1} \frac{3/5 + 1/4}{1 - (3/5 \cdot 1/4)} = \tan^{-1} \frac{17/20}{17/20} = \tan^{-1} 1$
$\therefore \frac{\pi}{4}$ (নির্ণেয় মান)।

খ-এর উত্তর:
দৃশ্যকল্প-১ এর চিত্র হতে, $\sin \theta = \frac{12}{13}, \cos \theta = \frac{5}{13}, \tan \theta = \frac{12}{5}$
$\implies \theta = \tan^{-1} \frac{12}{5}$
বামপক্ষ $= \frac{1}{2} \theta + \csc^{-1} \frac{5}{3} = \frac{1}{2} \tan^{-1} \frac{12}{5} + \tan^{-1} \frac{3}{4}$
ধরি, $\frac{1}{2} \tan^{-1} \frac{12}{5} = \alpha \implies \tan 2\alpha = \frac{12}{5}$
$\implies \frac{2\tan \alpha}{1 - \tan^{2} \alpha} = \frac{12}{5} \implies 10\tan \alpha = 12 - 12\tan^{2} \alpha$
$\implies 6\tan^{2} \alpha + 5\tan \alpha - 6 = 0 \implies (2\tan \alpha + 3)(3\tan \alpha - 2) = 0$
যেহেতু $\alpha$ সূক্ষ্মকোণ, $\tan \alpha = 2/3 \implies \alpha = \tan^{-1} \frac{2}{3}$
$\therefore \text{L.H.S} = \tan^{-1} \frac{2}{3} + \tan^{-1} \frac{3}{4} = \tan^{-1} \frac{2/3 + 3/4}{1 - 2/3 \cdot 3/4} = \tan^{-1} \frac{17/12}{6/12} = \tan^{-1} \frac{17}{6}$
ডানপক্ষ $= \cot^{-1} 2 + \cot^{-1} \frac{29}{28} = \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{28}{29}$
$\implies \tan^{-1} \frac{1/2 + 28/29}{1 - 1/2 \cdot 28/29} = \tan^{-1} \frac{85/58}{30/58} = \tan^{-1} \frac{17}{6}$
$\therefore \text{L.H.S = R.H.S}$ (প্রমাণিত)।

গ-এর উত্তর:
দেওয়া আছে, $f(x) = \cos x$
$\therefore \sqrt{3}\cos x + \cos(\frac{\pi}{2} + x) = 1 \implies \sqrt{3}\cos x - \sin x = 1$
উভয় পক্ষকে $\sqrt{(\sqrt{3})^{2} + (-1)^{2}} = 2$ দ্বারা ভাগ করে পাই,
$\frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x = \frac{1}{2} \implies \cos x \cos \frac{\pi}{6} - \sin x \sin \frac{\pi}{6} = \cos \frac{\pi}{3}$
$\implies \cos(x + \frac{\pi}{6}) = \cos \frac{\pi}{3} \implies x + \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{3}$
$\therefore x = 2n\pi + \frac{\pi}{6}$ অথবা $x = 2n\pi - \frac{\pi}{2}$
$-2\pi \le x < 2\pi$ ব্যবধিতে মানসমূহ:
$n = 0 \implies x = \frac{\pi}{6}, -\frac{\pi}{2}$
$n = 1 \implies x = \frac{13\pi}{6}$ (সীমার বাইরে), $x = \frac{3\pi}{2}$
$n = -1 \implies x = -\frac{11\pi}{6}, -\frac{5\pi}{2}$ (সীমার বাইরে)
$\therefore x = -\frac{11\pi}{6}, -\frac{\pi}{2}, \frac{\pi}{6}, \frac{3\pi}{2}$ (নির্ণেয় সমাধান)।