Question

1-tan²tita/1+tan²tita=cos²tita-sin²tita



please answer fast ​

Answer 1

Step-by-step explanation:

We have,

$\frac{1 - \tan^{2} ( \theta) }{1 + \tan^{2} ( \theta) } = \frac{1 - \tan^{2} ( \theta) }{ \sec^{2} ( \theta) }$

$\implies \frac{1 - \tan^{2} ( \theta) }{1 + \tan^{2} ( \theta) } = \frac{1 }{ \sec^{2} ( \theta) } - \frac{\tan^{2} ( \theta)}{\sec^{2} ( \theta)}$

$\implies \frac{1 - \tan^{2} ( \theta) }{1 + \tan^{2} ( \theta) } = \cos^{2} ( \theta) - \sin^{2} ( \theta)$