Question

answer this one immediately​

Answer 1

Step-by-step explanation:

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

hope it helps

Answer 2

Answer:

I think this will help you.