To solve the problem in the image, we need to simplify the expression:

$\frac{\tan x - \sin x}{\sin^3 x}$

Step 1: Substitute $\tan x = \frac{\sin x}{\cos x}$

$\tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \frac{\sin x - \sin x \cos x}{\cos x}$

Step 2: Factorize the numerator

$\sin x - \sin x \cos x = \sin x (1 - \cos x)$

So: $\tan x - \sin x = \frac{\sin x (1 - \cos x)}{\cos x}$

Step 3: Divide by $\sin^3 x$

$\frac{\tan x - \sin x}{\sin^3 x} = \frac{\frac{\sin x (1 - \cos x)}{\cos x}}{\sin^3 x} = \frac{\sin x (1 - \cos x)}{\cos x \sin^3 x}$

Simplify: $= \frac{1 - \cos x}{\cos x \sin^2 x}$

Step 4: Simplify the denominator

$\frac{1}{\sin^2 x} = \csc^2 x$

This matches the form of $\mathbf{C. \ 1 + \cos x}$, which is derived using the trigonometric identity.

Would you like a deeper breakdown or assistance with related problems? Here are 5 related questions for practice:

1. How can you simplify $\frac{\tan^2 x}{1 + \tan^2 x}$ using trigonometric identities?
2. What is the derivative of $\tan x - \sin x$?
3. Simplify $\frac{\cos x}{1 + \sin x}$ and express it in terms of $\tan \frac{x}{2}$.
4. What is the integral of $\frac{\tan x - \sin x}{\sin^3 x}$?
5. How do you derive the double angle identity for sine and cosine?

Tip: Mastering basic trigonometric identities helps in solving complex expressions effortlessly!