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ID#6604 HSC Higher Math 2nd MCQ (Rajshahi 2025)

$\sin^2(\cos^{-1}\frac{1}{\sqrt{3}})$ এর মান—
ক) $\frac{1}{\sqrt{3}}$
খ) $\frac{2}{\sqrt{3}}$
গ) $\frac{\sqrt{2}}{3}$
ঘ) $\frac{2}{3}$

ব্যাখ্যা

ধরা যাক $\theta = \cos^{-1}\frac{1}{\sqrt{3}}$। এর অর্থ হলো $\cos\theta = \frac{1}{\sqrt{3}}$। আমরা জানি যে $\sin^2\theta + \cos^2\theta = 1$। অতএব, $\sin^2\theta = 1 - \cos^2\theta$। এখানে $\cos\theta$ এর মান বসিয়ে পাই $\sin^2\theta = 1 - (\frac{1}{\sqrt{3}})^2 = 1 - \frac{1}{3} = \frac{2}{3}$।
Resource Details
Exam HSC
Subject Higher Math 2nd paper
Chapter 7
Board Rajshahi
Year 2025

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