Difficult problem below. Is the statement true or false?

Question

For all values for which the expressions are defined, sec x - tan x and

.. . . . . 1
__________
sec x + tan x

are equivalent expressions.

Answer 1

This is just an identity.

sec(x) - tan(x) = 1/[sec(x) + tan(x)]

Let's start with the right hand side and see if we can get the left hand side.

RHS = 1/[sec(x) + tan(x)]

Change everything to sines and cosines.

RHS = 1/[ 1/cos(x) + sin(x)/cos(x) ]

Multiply numerator and denominator by cos(x),

RHS = cos(x) / [1 + sin(x)]

Now, multiply numerator by the conjugate of the denominator.

RHS = cos(x) [1 - sin(x)] / ([1 + sin(x)][1 - sin(x)])
RHS = cos(x) [1 - sin(x)] / [1 - sin^2(x)]

Note that 1 - sin^2(x) = cos^2(x), so

RHS = cos(x) [1 - sin(x)] / cos^2(x)

Cancel terms,

RHS = [1 - sin(x)] / cos(x)

And now, make this into two fractions.

RHS = 1/cos(x) - sin(x)/cos(x)
RHS = sec(x) - tan(x) = LHS

It is true.

Answer 2

If the identity is true, then we must have:
sec^2x - tan^2 = 1
1/cos^2x- sin^2x/cos^2x = 1
(1-sin^2x)/cos^2x =1
1- sin^2x = cos^2x
1= sin^2x + cos^2x
1 = 1
The identity is true.