What is the value of $$\frac{1-\text{tan}^2\text{A}}{{1+\text{tan}^2\text{A}}}$$​ ?

Correct option is A

Given:

​$$\frac{1 - \tan^2 A}{1 + \tan^2 A}$$

​Formula Used:​

tan A = $$\frac{\sin A}{\cos A}$$ 

cos 2A = cos²A - sin²A

Solution:​

​$$\frac{1 - \tan^2 A}{1 + \tan^2 A}$$

= $$\frac{1-\frac{\sin^2A}{\cos^2A}}{1+\frac{\sin^2A}{\cos^2A}}$$

=$$\frac{\frac{\cos^2 A-\sin^2A}{\cos^2A}}{\frac{cos^2A+\sin^2A}{\cos^2A}}$$

= cos²A - sin²A

= cos 2A

​Important Points to Remember for Using Direct Approach in Difficult Questions:​

​ $$\text{sin } 2A = 2 \sin A \cos A = \frac{2 \tan A}{1 + \tan^2 A}\\\text{cos } 2A = \cos^2 A - \sin^2 A = 2 \cos^2 A - 1 = 1 - 2 \sin^2 A = \frac{1 - \tan^2 A}{1 + \tan^2 A}\\\text{tan } 2A = \frac{2 \tan A}{1 - \tan^2 A}\\$$​​