Question

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

Answer 1

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

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

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

= \(\frac{sec^2 \, \theta\, +\, tan^2 \, \theta \,+\,2\, sec\, \theta\, tan\,\theta}{1}\)      [sec² θ − tan² θ = 1]

= 1 + tan²θ + tan²θ + 2 secθ tanθ

= 1 + 2tan²θ + 2 secθ tanθ