Question

tan theta + sec theta minus 1 by tan theta + sec theta + 1 is equal to 1 + sin theta by cos theta prove it

Answer 1

Correction in Question :
-sec(a) in denominator.

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

2) LHS = $\frac{1+sin(\theta)}{cos(\theta)}$

Hence Proved