$\sec^2 \frac{14\pi}{17} - \cot^2 \frac{23\pi}{34}$ = কত?

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Accepted Answer

গ. 1

ব্যাখ্যা: আমরা জানি,
$\sec\left(\frac{14\pi}{17}ight) = \sec\left(\pi - \frac{3\pi}{17}ight) = -\sec\left(\frac{3\pi}{17}ight)$
অতএব, $\sec^2\left(\frac{14\pi}{17}ight) = \sec^2\left(\frac{3\pi}{17}ight)$
আবার,
$\cot\left(\frac{23\pi}{34}ight) = \cot\left(\frac{17\pi + 6\pi}{34}ight) = \cot\left(\frac{\pi}{2} + \frac{3\pi}{17}ight) = -	an\left(\frac{3\pi}{17}ight)$
অতএব, $\cot^2\left(\frac{23\pi}{34}ight) = 	an^2\left(\frac{3\pi}{17}ight)$
সুতরাং, $\sec^2\left(\frac{14\pi}{17}ight) - \cot^2\left(\frac{23\pi}{34}ight) = \sec^2\left(\frac{3\pi}{17}ight) - 	an^2\left(\frac{3\pi}{17}ight) = 1$।

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