sec⁡−1(x)=cos⁡−1(−12)+sin⁡−1(533) \sec^{-1}(x) = \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{5}{3\sqrt{3}}\right) সমীকরণটির সমাধান কোনটি?

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Accepted Answer

ঙ. −1815+6-\frac{18}{15+\sqrt{6}}

ব্যাখ্যা: দেওয়া আছে, $\sec^{-1}(x) = \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{5}{3\sqrt{3}}\right)$
ধরি, $\alpha = \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ এবং $\beta = \sin^{-1}\left(\frac{5}{3\sqrt{3}}\right)$
অতএব, $\cos\alpha = -\frac{1}{2}$, $\sin\alpha = \frac{\sqrt{3}}{2}$
এবং $\cos\beta = \sqrt{1 - \left(\frac{5}{3\sqrt{3}}\right)^2} = \sqrt{1 - \frac{25}{27}} = \sqrt{\frac{2}{27}} = \frac{\sqrt{2}}{3\sqrt{3}}$

এখন, $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$
$= \left(-\frac{1}{2}\right)\left(\frac{\sqrt{2}}{3\sqrt{3}}\right) - \left(\frac{\sqrt{3}}{2}\right)\left(\frac{5}{3\sqrt{3}}\right)$
$= -\frac{\sqrt{6}}{18} - \frac{15}{18} = -\frac{15 + \sqrt{6}}{18}$

সুতরাং, $x = \sec(\alpha + \beta) = \frac{1}{\cos(\alpha + \beta)} = -\frac{18}{15 + \sqrt{6}}$।

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