$cosec15^{\circ}$ এর মান কোনটি

$cosec15^{\circ} = \frac{1}{sin15^{\circ}}$ . আমরা জানি $sin(45^{\circ}-30^{\circ}) = sin45^{\circ}cos30^{\circ} - cos45^{\circ}sin30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$ . সুতরাং, $cosec15^{\circ} = \frac{2\sqrt{2}}{\sqrt{3}-1} = \frac{2\sqrt{2}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{2\sqrt{2}(\sqrt{3}+1)}{3-1} = \frac{2\sqrt{2}(\sqrt{3}+1)}{2} = \sqrt{2}(\sqrt{3}+1) = \sqrt{6}+\sqrt{2}$ . তবে, বিকল্পগুলিতে $\frac{4}{\sqrt{6}-\sqrt{2}}$ দেওয়া আছে। যদি $\frac{4}{\sqrt{6}-\sqrt{2}}$ কে সরল করি, তবে পাই $\frac{4(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} = \frac{4(\sqrt{6}+\sqrt{2})}{6-2} = \frac{4(\sqrt{6}+\sqrt{2})}{4} = \sqrt{6}+\sqrt{2}$ . অতএব, $cosec15^{\circ} = \sqrt{6}+\sqrt{2} = \frac{4}{\sqrt{6}-\sqrt{2}}$ .