Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 49

Differentiate each Trigonometric Identity to obtain new identity.
$\displaystyle \text{(a) }\tan x = \frac{\sin x}{\cos x} \qquad \text{(b) } \sec x = \frac{1}{\cos x} \qquad \text{(c) } \sin x + \cos x = \frac{1+ \cot x}{\csc x}$

$\begin{equation} \begin{aligned} \text{a.) } \tan x &= \frac{\sin x}{\cos x}\\ \\ \frac{d}{dx}(\tan x) &= \frac{\cos x \frac{d}{dx} \sin x - \sin x \frac{d}{dx} \cos x}{(\cos x)^2} && \text{Applying quotient rule}\\ \\ \sec ^2 x &= \frac{\cos x(\cos x) - \sin x ( - \sin x)}{ (\cos x)^2}\\ \\ \sec ^2 x &= \frac{\cos ^2 x + \sin ^2x}{(\cos x)^2} && \text{Applying the Pythagorean Identity for the trigonometric function } \sin^2x + \cos^2 x = 1\\ \\ \sec ^2 x &= \frac{1}{(\cos x)^2} && \text{Applying the identity } \sec x = \frac{1}{\cos x}\\ \\ \sec ^2 x & = \sec ^2 x \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{b.) } \sec x &= \frac{1}{\cos x}\\ \\ \frac{d}{dx} \sec x & = \frac{\cos x \frac{d}{dx} (1) - 1 \frac{d}{dx} \cos x}{(\cos x)^2} && \text{Applying quotient rule}\\ \\ \sec x \tan x & = \frac{0-1 ( - \sin x ) }{(\cos x)^2}\\ \\ \sec x \tan x & = \frac{\sin x}{(\cos x)^2} && \text{Break down two parts}\\ \\ \sec x \tan x & = \frac{\sin x}{\cos x} \cdot \left( \frac{1}{\cos x}\right)\\ \\ \sec x \tan x & = \sec x \tan x \end{aligned} \end{equation}$



$\begin{equation} \begin{aligned} \text{c.) } \sin x + \cos x &= \frac{1+ \cot x}{\csc x}\\ \\ \frac{d}{dx} \sin x + \frac{d}{dx} \cos x &= \frac{\csc x \frac{d}{dx} (1+\cot x) - (1+\cot x) \frac{d}{dx} \csc x}{(\csc x)^2}\\ \\ \cos x - \sin x &= \frac{\csc x (-\csc^2 x) - ( 1+ \cot x) (-\csc x \cot x)}{(\csc x)^2}\\ \\ \cos x - \sin x &= \frac{-\csc ^3 x + \csc x \cot x + \csc x \cot^2 x}{(\csc x)^2}\\ \\ \cos x - \sin x &= \frac{-\csc^3 x + \csc x \frac{\cos x}{\sin x} + \csc x \frac{\cos^2 x}{\sin ^2 x}}{(\csc x)^2}\\ \\ \cos x - \sin x &= \frac{-\csc^3 x + \csc^2 x \cos x + \csc^3 x \cos x}{(\csc x)^2}\\ \\ \cos x - \sin x &= \frac{\cancel{\csc ^2 x}(- \csc x + \cos x + \cos x \csc x)}{\cancel{(\csc x)^2}}\\ \\ \cos x - \sin x &= \frac{1}{\sin x} + \cos x + \frac{\cos^2 x}{\sin x}\\ \\ \cos x - \sin x &= \frac{\cos ^2 x - 1}{ \sin x} + \cos x && (\cos^x - 1 = -\sin^2 x, \text{ Using Pythagorean Identity})\\ \\ \cos x - \sin x &= - \frac{\sin ^{\cancel{2}}x}{\cancel{\sin x}} + \cos x\\ \end{aligned} \end{equation}\\ \qquad \boxed{\cos x - \sin x = \cos x - \sin x }$