nework Prove the identity. \[ \frac{1-\csc x}{\csc x+1}-\frac{\csc x+1}{1-\csc x}=4 \sec x \tan x \] To verify the identity, work each side separately until you obtain the same expression. Start with the left side. Choose the correct step and transform the expression according to the step chosen. 15 \( \frac{1-\csc x}{\csc x+1}-\frac{\csc x+1}{1-\csc x} \) \( =\square \)

We wish to show that   (1 – csc x)/(csc x + 1) – (csc x + 1)/(1 – csc x) = 4 sec x tan x. One very neat choice for a first step is to combine the two fractions over a common denominator. That is, write   (1 – csc x)/(csc x + 1) – (csc x + 1)/(1 – csc x)     = { (1 – csc x)² – (csc x + 1)² }⁄[ (csc x + 1)(1 – csc x) ]. This is the correct step that “transforms the expression according to the step chosen” (i.e. choosing to combine the fractions by writing a common denominator). Let’s now quickly see how this step leads us to the answer. Step 1. Write as a single fraction:   (1 – csc x)/(csc x + 1) – (csc x + 1)/(1 – csc x)     = [ (1 – csc x)² – (csc x + 1)² ]⁄[ (csc x + 1)(1 – csc x) ]. Step 2. Notice that the numerator is a difference of two squares:   (1 – csc x)² – (csc x + 1)² = [(1 – csc x) – (csc x + 1)]·[(1 – csc x) + (csc x + 1)]. Compute each factor:   First factor: (1 – csc x) – (csc x + 1) = 1 – csc x – csc x – 1 = –2 csc x.   Second factor: (1 – csc x) + (csc x + 1) = 1 – csc x + csc x + 1 = 2. Thus, the numerator becomes:   (–2 csc x)(2) = –4 csc x. Step 3. Notice that the denominator simplifies by using the difference of two factors:   (csc x + 1)(1 – csc x) = 1 – csc² x. So the expression is now:   –4 csc x/(1 – csc² x). Step 4. Recall the Pythagorean identity for cosecant:   csc² x = 1 + cot² x  or equivalently 1 – csc² x = –cot² x. While this is one route, an even simpler method is to write everything in terms of sine and cosine. Using the definitions:   csc x = 1/sin x,  sec x = 1/cos x,  tan x = sin x/cos x, we can proceed by rewriting from the very beginning. Alternate route using sine and cosine:   Write LHS as   (1 – 1/sin x)/(1/sin x + 1) – (1/sin x + 1)/(1 – 1/sin x). Multiply numerator and denominator in each term by sin x:   = (sin x – 1)/(1 + sin x) – (1 + sin x)/(sin x – 1). Now observe that the two fractions are negatives of one another. Writing them over a common denominator:   = { (sin x – 1)² – (sin x + 1)² }⁄[ (sin x + 1)(sin x – 1) ]. The numerator becomes:   (sin x – 1)² – (sin x + 1)² = [sin²x – 2sin x + 1] – [sin²x + 2sin x + 1] = –4 sin x. And the denominator is   (sin x + 1)(sin x – 1) = sin²x – 1 = –(1 – sin²x) = –cos² x. Thus,   LHS = (–4 sin x)/(–cos² x) = 4 sin x/cos² x = 4 (sin x/cos x)(1/cos x) = 4 tan x sec x,  which is the desired right side. To summarize, one acceptable starting step is to rewrite the left side as   (1 – csc x)/(csc x + 1) – (csc x + 1)/(1 – csc x)     = [(1 – csc x)² – (csc x + 1)²]⁄[ (csc x + 1)(1 – csc x) ]. This is the step that transforms the expression into one that can then be simplified to obtain 4 sec x tan x.