ID#6152 HSC Higher Math 2nd CQ (Barisal 2025)

ক-এর উত্তর:
বামপক্ষ = $\tan^{-1} \frac{2}{9} + \tan^{-1} \frac{1}{4}$
$= \tan^{-1} \frac{\frac{2}{9} + \frac{1}{4}}{1 - \frac{2}{9} \cdot \frac{1}{4}}$
$= \tan^{-1} \frac{\frac{8+9}{36}}{\frac{36-2}{36}}$
$= \tan^{-1} \frac{17}{34} = \tan^{-1} \frac{1}{2}$

ধরি, $\tan^{-1} \frac{1}{2} = \theta \implies \tan \theta = \frac{1}{2}$
আমরা জানি, $\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^{2} \theta} = \frac{2 \cdot \frac{1}{2}}{1 + (\frac{1}{2})^{2}} = \frac{1}{1 + \frac{1}{4}} = \frac{4}{5}$
$\therefore 2\theta = \sin^{-1} \frac{4}{5} \implies \theta = \frac{1}{2} \sin^{-1} \frac{4}{5}$
$\therefore \tan^{-1} \frac{1}{2} = \frac{1}{2} \sin^{-1} \frac{4}{5}$ = ডানপক্ষ (প্রমাণিত)

খ-এর উত্তর:
দেওয়া আছে, $f(x) = \cos x$
$\therefore f(\frac{\pi}{2} - \theta) = \cos(\frac{\pi}{2} - \theta) = \sin \theta$ এবং $f(2\theta) = \cos 2\theta$

বামপক্ষ = $\sin^{-1} \sqrt{2 \sin \theta} + \sin^{-1} \sqrt{\cos 2\theta}$
ধরি, $\sin^{-1} \sqrt{2 \sin \theta} = A \implies \sin^{2} A = 2 \sin \theta$ --- (i)
এবং $\sin^{-1} \sqrt{\cos 2\theta} = B \implies \sin^{2} B = \cos 2\theta = 1 - 2 \sin^{2} \theta$ --- (ii)

আদর্শ প্রশ্ন কাঠামো অনুযায়ী যদি প্রথম পদটি $\sin^{-1} (\sqrt{2} \sin \theta)$ হতো, তবে $\sin^2 A = 2 \sin^2 \theta$ হতো।
তখন $\sin^2 A + \sin^2 B = 2 \sin^2 \theta + 1 - 2 \sin^2 \theta = 1$
$\therefore \sin^2 A = 1 - \sin^2 B = \cos^2 B \implies \sin A = \cos B \implies A = \frac{\pi}{2} - B$
$\therefore A + B = \frac{\pi}{2}$ (প্রমাণিত)

গ-এর উত্তর:
দেওয়া আছে, $\sqrt{2} \cos \alpha - \sqrt{2} \cos(\frac{\pi}{2} - \alpha) = 1$
বা, $\sqrt{2} \cos \alpha - \sqrt{2} \sin \alpha = 1$
বা, $\cos \alpha - \sin \alpha = \frac{1}{\sqrt{2}}$
বা, $\frac{1}{\sqrt{2}} \cos \alpha - \frac{1}{\sqrt{2}} \sin \alpha = \frac{1}{2}$ [$\sqrt{2}$ দ্বারা ভাগ করে]
বা, $\cos \frac{\pi}{4} \cos \alpha - \sin \frac{\pi}{4} \sin \alpha = \frac{1}{2}$
বা, $\cos(\alpha + \frac{\pi}{4}) = \cos \frac{\pi}{3}$

$\therefore \alpha + \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{3}$
বা, $\alpha = 2n\pi \pm \frac{\pi}{3} - \frac{\pi}{4}$

$n=0$ হলে, $\alpha = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$ এবং $\alpha = -\frac{\pi}{3} - \frac{\pi}{4} = -\frac{7\pi}{12}$
প্রদত্ত সীমার মধ্যে ($-\pi < \alpha < \pi$) নির্ণেয় সমাধান: $\alpha = \frac{\pi}{12}, -\frac{7\pi}{12}$