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

$\sec^2\frac{14\pi}{17} - \cot^2\frac{23\pi}{34}$ আমরা জানি, $\cot x = \tan (\frac{\pi}{2} - x)$ . তাই, $\cot\frac{23\pi}{34} = \tan(\frac{\pi}{2} - \frac{23\pi}{34}) = \tan(\frac{17\pi - 23\pi}{34}) = \tan(\frac{-6\pi}{34}) = \tan(\frac{-3\pi}{17}) = -\tan(\frac{3\pi}{17})$ . আবার, $\sec^2 x = 1 + \tan^2 x$ . সুতরাং, $\sec^2\frac{14\pi}{17} = 1 + \tan^2\frac{14\pi}{17}$ . এবং $\cot^2\frac{23\pi}{34} = (-\tan\frac{3\pi}{17})^2 = \tan^2\frac{3\pi}{17}$ . আমরা জানি, $\tan(\pi - x) = -\tan x$ . সুতরাং, $\tan\frac{14\pi}{17} = \tan(\pi - \frac{3\pi}{17}) = -\tan\frac{3\pi}{17}$ . তাই, $\tan^2\frac{14\pi}{17} = (-\tan\frac{3\pi}{17})^2 = \tan^2\frac{3\pi}{17}$ . অতএব, $\sec^2\frac{14\pi}{17} - \cot^2\frac{23\pi}{34} = (1 + \tan^2\frac{14\pi}{17}) - \tan^2\frac{3\pi}{17}$ . যেহেতু $\tan^2\frac{14\pi}{17} = \tan^2\frac{3\pi}{17}$ , তাই $1 + \tan^2\frac{14\pi}{17} - \tan^2\frac{3\pi}{17} = 1 + \tan^2\frac{3\pi}{17} - \tan^2\frac{3\pi}{17} = 1$ .