# Question #cf786

Mar 2, 2017

see below

#### Explanation:

Use the following Properties:

1. Pythagorean Properties

$1 + {\cot}^{2} x = {\csc}^{2} x$

$1 + {\tan}^{2} x = {\sec}^{2} x$

${\cos}^{2} x + {\sin}^{2} x = 1$

2. Reciprocal Properties
$\csc x = \frac{1}{\sin} x$

$\sec x = \frac{1}{\cos} x$

Left hand Side :

${\left(1 - \cot x\right)}^{2} + {\left(1 - \tan x\right)}^{2} = 1 - 2 \cot x + {\cot}^{2} x + 1 - 2 {\tan}^{2} x + {\tan}^{2} x$

$= 1 + {\cot}^{2} x - 2 \cot x - 2 \tan x + 1 + {\tan}^{2} x$

$= {\csc}^{2} x - 2 \left(\cot x + \tan x\right) + {\sec}^{2} x$

$= {\csc}^{2} x - 2 \left(\cos \frac{x}{\sin} x + \sin \frac{x}{\cos} x\right) + {\sec}^{2} x$

$= {\csc}^{2} x - 2 \left(\frac{{\cos}^{2} x + {\sin}^{2} x}{\sin x \cos x}\right) + {\sec}^{2} x$

$= {\csc}^{2} x - 2 \left(\frac{1}{\sin x \cos x}\right) + {\sec}^{2} x$

$= {\csc}^{2} x - 2 \left(\frac{1}{\sin} x \frac{1}{\cos} x\right) + {\sec}^{2} x$

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

$= {\csc}^{2} x - 2 \csc x \sec x + {\sec}^{2} x$---> factor

$= \left(\csc x - \sec x\right) \left(\csc x - \sec x\right)$

$= {\left(\csc x - \sec x\right)}^{2}$

$\therefore =$ Right Hand Side