Question

1/secx-tanx + 1/secx+tanx = 2/cosx

Answer 1

=1/sec x-tan x + 1/sec x+tan x
=Taking lcm
=Sec x + tan x+ sec x-tan x
/Sec x+ tan x * sec x - tan x
=2secx/secx.secx - tan x.tan x
=2sec x/1
=2/cos x
=RHS.

Answer 2

Another method

LHS

(1)/(1/cos x  -  sin x/cos x) + (1)/(1/cos x  +  sin x/cos x)

(1)/(1 - sin x / cos x) + (1)/(1 + sin x / cos x)

(cos x / 1-sin x) + (cos x / 1+sin x)

(cos x + cos x . sin x + cos x - cos x . sin x)/(1^2 - (sin x)^2)  {Cross Multiplication}

(2 cos x) / (1 - sin^2 x)

(2 cos x) / (cos^2 x)

(2 . cos x) / (cos x . cos x)

2 / cos x     = RHS

LHS = RHS

Hence proved

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