# How do you prove \frac { \tan x } { \sec x } + \frac { \cot x } { \csc x } = \sin x + \cos x?

Dec 5, 2016

See the explanation

#### Explanation:

$\frac{\textcolor{red}{\tan x}}{\textcolor{b l u e}{\sec x}} + \frac{\textcolor{g r e e n}{\cot x}}{\textcolor{m a \ge n t a}{\csc x}} = \sin x + \cos x$

Rewrite $\textcolor{red}{\tan} x$ as $\frac{\textcolor{b r o w n}{\sin x}}{\textcolor{D a r k T u r q u o i s e}{\cos x}}$

Rewrite $\textcolor{b l u e}{\sec} x$ as $\frac{1}{\textcolor{D a r k T u r q u o i s e}{\cos x}}$

Rewrite $\textcolor{g r e e n}{\cot} x$ as $\frac{1}{\textcolor{red}{\tan x}}$

Rewrite $\textcolor{m a \ge n t a}{\csc} x$ as $\frac{1}{\textcolor{b r o w n}{\sin x}}$

$\frac{\frac{\textcolor{b r o w n}{\sin x}}{\textcolor{D a r k T u r q u o i s e}{\cos x}}}{\frac{1}{\textcolor{D a r k T u r q u o i s e}{\cos x}}} + \frac{\frac{1}{\textcolor{red}{\tan x}}}{\frac{1}{\textcolor{b r o w n}{\sin x}}} = \sin x + \cos x$

$\frac{\textcolor{b r o w n}{\sin x} \cancel{\textcolor{D a r k T u r q u o i s e}{\cos x}}}{\cancel{\textcolor{D a r k T u r q u o i s e}{\cos x}}} + \frac{\textcolor{b r o w n}{\sin x}}{\textcolor{red}{\tan x}} = \sin x + \cos x$

Rewrite $\textcolor{red}{\tan} x$ as $\frac{\textcolor{b r o w n}{\sin x}}{\textcolor{D a r k T u r q u o i s e}{\cos x}}$

$\textcolor{b r o w n}{\sin} x + \frac{\textcolor{b r o w n}{\sin x}}{\frac{\textcolor{b r o w n}{\sin x}}{\textcolor{D a r k T u r q u o i s e}{\cos x}}} = \sin x + \cos x$

$\textcolor{b r o w n}{\sin} x + \frac{\cancel{\textcolor{b r o w n}{\sin x}} \textcolor{D a r k T u r q u o i s e}{\cos x}}{\cancel{\textcolor{b r o w n}{\sin}}} x = \sin x + \cos x$

$\textcolor{b r o w n}{\sin} x + \textcolor{D a r k T u r q u o i s e}{\cos} x = \sin x + \cos x$