(cos27°-cos63°)/(cos27°+cos63°)=? 

Explanation:

Another Explanation (5): গণিত 🧮 সমস্যা: \(\frac{\cos 27^\circ - \cos 63^\circ}{\cos 27^\circ + \cos 63^\circ}\) = ? আমরা জানি, \(\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}\) এবং \(\cos C + \cos D = 2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}\) সুতরাং, \(\frac{\cos 27^\circ - \cos 63^\circ}{\cos 27^\circ + \cos 63^\circ} = \frac{-2 \sin \frac{27^\circ + 63^\circ}{2} \sin \frac{27^\circ - 63^\circ}{2}}{2 \cos \frac{27^\circ + 63^\circ}{2} \cos \frac{27^\circ - 63^\circ}{2}}\) \(= \frac{-2 \sin \frac{90^\circ}{2} \sin \frac{-36^\circ}{2}}{2 \cos \frac{90^\circ}{2} \cos \frac{-36^\circ}{2}}\) \(= \frac{-2 \sin 45^\circ \sin (-18^\circ)}{2 \cos 45^\circ \cos (-18^\circ)}\) আমরা জানি, \(\sin(-x) = -\sin(x)\) এবং \(\cos(-x) = \cos(x)\) সুতরাং, \(= \frac{-2 \sin 45^\circ (-\sin 18^\circ)}{2 \cos 45^\circ \cos 18^\circ}\) \(= \frac{2 \sin 45^\circ \sin 18^\circ}{2 \cos 45^\circ \cos 18^\circ}\) \(= \frac{\sin 45^\circ}{\cos 45^\circ} \cdot \frac{\sin 18^\circ}{\cos 18^\circ}\) আমরা জানি, \(\frac{\sin x}{\cos x} = \tan x\) সুতরাং, \(= \tan 45^\circ \cdot \tan 18^\circ\) আমরা জানি, \(\tan 45^\circ = 1\) সুতরাং, \(= 1 \cdot \tan 18^\circ\) \(= \tan 18^\circ\) অতএব, \(\frac{\cos 27^\circ - \cos 63^\circ}{\cos 27^\circ + \cos 63^\circ} = \tan 18^\circ\) 🎉