(1-tan^2 (45^o +2x))/(1+tan^2 (45^o +2x) = ?

Another Explanation (5):

সমাধান:

প্রশ্ন: \(\frac{1 - \tan^2 (45^\circ + 2x)}{1 + \tan^2 (45^\circ + 2x)} = ?\) প্রথমে, আমরা মনে রাখি যে, \[ \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta \] এখানে, \(\theta = 45^\circ + 2x\) অতএব, \[ \frac{1 - \tan^2 (45^\circ + 2x)}{1 + \tan^2 (45^\circ + 2x)} = \cos 2 (45^\circ + 2x) \] এখন, \[ 2 (45^\circ + 2x) = 2 \times 45^\circ + 2 \times 2x = 90^\circ + 4x \] তাই, \[ \frac{1 - \tan^2 (45^\circ + 2x)}{1 + \tan^2 (45^\circ + 2x)} = \cos (90^\circ + 4x) \] আমরা জানি, \(\cos (90^\circ + \theta) = - \sin \theta\) অতএব, \[ \cos (90^\circ + 4x) = - \sin 4x \] সুতরাং, উত্তর হলো:

উত্তর: \(- \sin 4x\)