Prove : (tan θ)/(sec θ) = (tan θ + sec θ + 1)/(tan θ + sec θ - 1)

Question

Prove :

\(\frac{\tan \theta}{sec\, \theta} = \frac{tan\,\theta + sec\,\theta + 1}{tan\,\theta + sec\,\theta - 1}\)

(tan θ)/(sec θ) = (tan θ + sec θ + 1)/(tan θ + sec θ - 1)

Answer 1

tan θ/(sec θ - 1)

\(\therefore\) By theorem on equal ratios,