cos(sec^-1(x^2/y^2)) × sec(cos(y^2/x^2))  = কত ?

Another Explanation (5): প্রশ্ন: \(\cos(\sec^{-1}(\frac{x^2}{y^2})) \times \sec(\cos(\frac{y^2}{x^2})) = কত ?\) উত্তর: 1 সমাধান: প্রথম অংশ: \(\cos(\sec^{-1}(\frac{x^2}{y^2}))\) ধরি, \(\theta = \sec^{-1}\left(\frac{x^2}{y^2}\right)\) তাহলে, \(\sec \theta = \frac{x^2}{y^2}\) এবং, \(\sec \theta = \frac{\textউঁচু, \(\cos \theta = \frac{1}{\sec \theta} = \frac{y^2}{x^2}\) অতএব, \[ \cos(\sec^{-1}(\frac{x^2}{y^2})) = \frac{y^2}{x^2} \] --- দ্বিতীয় অংশ: \(\sec(\cos(\frac{y^2}{x^2}))\) ধরি, \(\phi = \cos^{-1}\left(\frac{y^2}{x^2}\right)\) তাহলে, \[ \cos \phi = \frac{y^2}{x^2} \] এবং, \[ \sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - \left(\frac{y^2}{x^2}\right)^2} = \sqrt{1 - \frac{y^4}{x^4}} = \frac{\sqrt{x^4 - y^4}}{x^2} \] সুতরাং, \[ \sec \phi = \frac{1}{\cos \phi} = \frac{x^2}{y^2} \] তাহলে, \[ \sec(\cos^{-1}(\frac{y^2}{x^2})) = \frac{x^2}{y^2} \] --- অতএব, মূল সমীকরণ: \[ \cos(\sec^{-1}(\frac{x^2}{y^2})) \times \sec(\cos(\frac{y^2}{x^2})) = \left(\frac{y^2}{x^2}\right) \times \left(\frac{x^2}{y^2}\right) = 1 \] **উত্তর: \(\boxed{1}\)**