Question

Prove that :
1/( sec x - tan x) - 1/cos x = 1/ cos x - 1/( sec x + tan x).

Give give detailed steps. It's a request.​

Answer 1

Step-by-step explanation:

$Given:\\\frac{1}{sec \: x - tan \: x} - \frac{1}{cos \: x} \\ = \frac{1}{cos \: x} - \frac{1}{sec \: x + tan \: x} \\ \\ \therefore \: \frac{1}{sec \: x - tan \: x} + \frac{1}{sec \: x + tan \: x} \\ \\ = \frac{1}{cos \: x} + \frac{1}{cos \: x} = \frac{2}{cos \: x} = 2 \: sec \: x \\ \\ LHS \\= \frac{1}{sec \: x - tan \: x} + \frac{1}{sec \: x + tan \: x} \\ \\ = \frac{sec \: x + tan \: x + sec \: x - tan \: x}{sec ^{2} \: x - tan^{2} \: x} \\ \\ = \frac{2 \: sec \: x }{1 }.. (\because sec ^{2} \: x - tan^{2} \: x=1) \\ \\ = 2 \: sec \: x \\ \\ = RHS \\ \\ Thus \: Proved$