Here we need to prove that cos C / sin (90 - C) - 1 = 1 - cos B / sin(90 - B)

Use the relation sin (90 - x) = cos x and cos (90 - x) = sin x

Let's start with the left hand side

cos C / sin (90 - C) - 1

=> cos C / cos C - 1

=> 1 - 1

=> 0

Now the right hand side

1 - cos B / sin(90 - B)

=> 1 - cos B / cos B

=> 1 - 1

=> 0

As the two sides equal 0 they are equal.**This proves that cos C / sin (90 - C) - 1 = 1 - cos B / sin(90 - B)**

** **

We'll re-write the identity, shifting it's terms:

cosC/sin(90-C) + cosB/sin(90-B) = 1+1

cosC/sin(90-C) + cosB/sin(90-B) = 2

We'll calculate the first ratio: cosB/sinC

We know that in a right angle triangle, we'll have the identities:

B = pi/2 - C

Now, we'll apply cosine function both sideS:

cos B = cos (pi/2 - C)

cos B = cos pi/2*cos C + sin pi/2*sin C

cos pi/2 = 0 and sin pi/2 =1

cos B = sin C

cosB/sinC = sin C/sin C = 1

We'll apply the same identities for the other fraction:

cosC/sinB = cos (90 - B)/sin B

cosC/sinB = sin B/sin B

cosC/sinB = 1

Managing the left side, we'll get:

LHS = 1 + 1 = 2 = RHS

**We notice that we've get the same result both sides, so the identity cosC/sin(90-C)-1=1-cosB/sin(90-B) is true.**