# Question #cab60

Apr 15, 2016

Recall the following identities:

$1. \textcolor{red}{\sec x = \frac{1}{\cos} x}$

$2. \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\tan x = \sin \frac{x}{\cos} x}$

$3. \textcolor{b l u e}{\cot x = \cos \frac{x}{\sin} x}$

$4. \textcolor{p u r p \le}{{\sin}^{2} x + {\cos}^{2} x = 1}$

Given the following identity, start the proof by working on the left side.

$\sec \frac{x}{\cos} x - \tan \frac{x}{\cot} x = 1$

Left side:

$\frac{\textcolor{red}{\frac{1}{\cos} x}}{\cos} x - \frac{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\sin \frac{x}{\cos} x}}{\textcolor{b l u e}{\cos \frac{x}{\sin} x}}$

$= \frac{1}{\cos} x \cdot \frac{1}{\cos} x - \sin \frac{x}{\cos} x \cdot \sin \frac{x}{\cos} x$

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

$= \frac{\textcolor{p u r p \le}{1 - {\sin}^{2} x}}{\cos} ^ 2 x$

$= {\cos}^{2} \frac{x}{\cos} ^ 2 x$

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 1 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\therefore$, left side$=$right side.

Apr 15, 2016

$\sec \frac{x}{\cos} x - \tan \frac{x}{\cot} x$

$= {\sec}^{2} x - {\tan}^{2} x$

$= 1$

#### Explanation:

LHS: $\sec \frac{x}{\cos} x - \tan \frac{x}{\cot} x$

$= \sec x \times \frac{1}{\cos} x - \tan x \times \frac{1}{\cot} x$

$= \sec x \times \sec x - \tan x \times \tan x$

$= {\sec}^{2} x - {\tan}^{2} x$

$= 1$