Simplify [sin (x - pi/2)) / cos (pi - x)] + [tan (x - 3pi/2) / - tan (pi + x)

We'll re-write sin (x - pi/2) as -sin (pi/2 - x) = - cos
x.

cos (pi - x) = cos pi*cos x + sin pi*sin
x

But sin pi = 0 and cos pi =
-1

cos (pi - x) = -cos x

The
first term of the sum will become:

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

We'll re-write the numerator
of the second term:

tan (x - 3pi/2) = - tan
x

The denominator of the 2nd term
is:

-tan(pi+x) = -(tan pi + tan x)/(1 - tanpi*tan
x)

But tan pi = 0

-tan(pi+x) =
-tan x/1

-tan(pi+x) = -tan
x

We'll re-write the second
term:

[tan (x - 3pi/2) / - tan (pi + x) = -tan x/-tan x =
1

The sum of the terms will
be:

[sin (x - pi/2)) / cos (pi - x)] + [tan (x - 3pi/2) / -
tan (pi + x) = 1+1

[sin (x - pi/2)) / cos (pi
- x)] + [tan (x - 3pi/2) / - tan (pi + x) = 2