1/ sec x - tan x -1/cos x= 1/ cos x- 1/ sec x+ tan x

1.	1/ sec x - tan x -1/cos x= 1/ cos x- 1/ sec x+ tan x
Answer» LHS =\xa0{tex}\\frac{1}{sec x - tan x}{/tex}\xa0-\xa0{tex}\\frac{1}{cos x}{/tex}={tex}\\frac{sec x + tan x}{(sec x - tan x)(sec x + tan x)}{/tex}\xa0-\xa0{tex}\\frac{1}{cos x}{/tex}=\xa0{tex}\\frac{secx + tan x}{sec^2 x - tan^2 x}{/tex}\xa0-\xa0{tex}\\frac{1}{cos x}{/tex}{tex}=sec x + tan x - sec x{/tex}{tex}[\\because sec^2\\theta-tan^2\\theta=1]{/tex}= tan xRHS =\xa0{tex}\\frac{1}{cosx - 1}{/tex}\xa0-\xa0{tex}\\frac{1}{sec x + tan x}{/tex}={tex}\\frac{1}{cos x}{/tex}\xa0-\xa0{tex}\\frac{sec x - tan x}{(sec x + tan x)(sec x - tan x)}{/tex}={tex}\\frac{1}{cos x}{/tex}\xa0-\xa0{tex}\\frac{sec x - tan x}{sec^2 x - tan^2 x}{/tex}= sec x -\xa0{tex}\\frac{sec x - tan x}{1}{/tex}{tex}=sec x - sec x + tan x{/tex}{tex}=tan x{/tex}Hence, LHS = RHS

Discussion