1. **State the problem:** We need to verify the identity $$\frac{\sin x + \cos x}{\cos^3 x} = 1 + \tan x + \tan^2 x + \tan^3 x$$. 2. **Recall definitions and formulas:** - $\tan x = \frac{\sin x}{\cos x}$. - We will express the left side in terms of $\tan x$ to compare with the right side. 3. **Rewrite the left side:** $$\frac{\sin x + \cos x}{\cos^3 x} = \frac{\sin x}{\cos^3 x} + \frac{\cos x}{\cos^3 x} = \frac{\sin x}{\cos^3 x} + \frac{1}{\cos^2 x}$$ 4. **Express in terms of $\tan x$:** - $\frac{\sin x}{\cos^3 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos^2 x} = \tan x \sec^2 x$ - $\frac{1}{\cos^2 x} = \sec^2 x$ So the left side becomes: $$\tan x \sec^2 x + \sec^2 x = \sec^2 x (1 + \tan x)$$ 5. **Recall the Pythagorean identity:** $$\sec^2 x = 1 + \tan^2 x$$ 6. **Substitute $\sec^2 x$ back:** $$\sec^2 x (1 + \tan x) = (1 + \tan^2 x)(1 + \tan x)$$ 7. **Expand the right side:** $$ (1 + \tan^2 x)(1 + \tan x) = 1(1 + \tan x) + \tan^2 x (1 + \tan x) = 1 + \tan x + \tan^2 x + \tan^3 x$$ 8. **Conclusion:** The left side simplifies exactly to the right side, so the identity is true. **Final answer:** $$\frac{\sin x + \cos x}{\cos^3 x} = 1 + \tan x + \tan^2 x + \tan^3 x$$