Question

1+sec0 - tan0 by 1 + sec0 + tan0 = 1 - sin0 by cos0​

Answer 1

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