\frac { 1 } { \operatorname { sec } x - \operatorname { tan } x } - \f

1.	\frac { 1 } { \operatorname { sec } x - \operatorname { tan } x } - \frac { 1 } { \operatorname { cos } x } = \frac { 1 } { \operatorname { cos } x } - \frac { 1 } { \operatorname { sec } x + \operatorname { tan } x }
Answer» 1/(sec x - tan x) - 1/cos x = 1/cos x - 1/(sec x + tan x) 1/sec x - tanx + 1/sec x + tan x= 2/cos x LHS:1/sec x - tanx + 1/sec x + tan x=sec x + tanx + sec x - tan x/ sec ^2 c + tan^2 x= 2sec x /1= 2/cos x= RHS Hence proved

\frac { 1 } { \operatorname { sec } x - \operatorname { tan } x } - \frac { 1 } { \operatorname { cos } x } = \frac { 1 } { \operatorname { cos } x } - \frac { 1 } { \operatorname { sec } x + \operatorname { tan } x }

Answer»

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

1/sec x - tanx + 1/sec x + tan x= 2/cos x

LHS:1/sec x - tanx + 1/sec x + tan x=sec x + tanx + sec x - tan x/ sec ^2 c + tan^2 x= 2sec x /1= 2/cos x= RHS

Hence proved