We have to prove \(\frac{1}{secθ − tanθ}\) – \(\frac{1}{cosθ}\) = \(\frac{1}{cosθ}\) – \(\frac{1}{secθ + tanθ}\) .

⇒ \(\frac{1}{secθ − tanθ}\) + \(\frac{1}{secθ + tanθ}\) = \(\frac{1}{cosθ}\) + \(\frac{1}{cosθ}\)

 ⇒ \(\frac{1}{secθ − tanθ}\)  + \(\frac{1}{secθ + tanθ}\) = \(\frac{2}{cosθ}\).

To proving \(\frac{1}{secθ − tanθ}\) - \(\frac{1}{cosθ}\) = \(\frac{1}{cosθ}\) - \(\frac{1}{secθ + tanθ}\) 

we have required to prove \(\frac{1}{secθ + tanθ}\) + \(\frac{1}{secθ + tanθ}\) = \(\frac{2}{cosθ}\).

Now, \(\frac{1}{secθ − tanθ}\) + \(\frac{1}{secθ + tanθ}\) = \(\frac{(sec\theta + tan \theta)(sec\theta -tan\theta)}{(sec\theta-tan\theta)(sec\theta +tan \theta}\) = \(\frac{2sec\theta}{sec^2 \theta - \tan^2\theta}\)

(∵ (a + b)(a – b) =a² − b² )

= 2 secθ = \(\frac{2}{cos \theta}\). (∵ sec²θ − tan²θ = 1 & secθ =\(\frac{1}{cosθ}\))

Hence, \(\frac{1}{sec\theta - tan \theta} + \frac{1}{sec\theta + tan \theta}\) = \(\frac{2}{cos\theta}\) = \(\frac{1}{cos\theta} + \frac{1}{cos\theta}\)

⇒ \(\frac{1}{sec\theta-tan \theta} - \frac{1}{cos \theta}\) = \(\frac{1}{cos \theta}\) - \(\frac{1}{sec\theta +tan \theta}\).

Hence proved