# How do you prove 1/(sinxcosx) - cosx/sinx = tanx?

Jun 26, 2018

Kindly see a Proof in Explanation.

#### Explanation:

$\frac{1}{\sin x \cos x} - \cos \frac{x}{\sin} x$,

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

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

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

$= \sin \frac{x}{\cos} x$,

$= \tan x$, as desired!

BONUS :

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

$\therefore \frac{1}{\sin x \cos x} - \cos \frac{x}{\sin} x = \tan x \Rightarrow 2 \csc 2 x - \cot x = \tan x$.

Jun 26, 2018

As proved below.

#### Explanation:

$\text{To prove } \frac{1}{\sin x \cos x} - \cos \frac{x}{\sin} x = \tan x$

$L H S = \frac{1}{\sin x \cos x} - \cos \frac{x}{\sin} x$

$\implies \frac{1 - \cos x \cdot \cos x}{\sin x \cos x} , \text{ taking " sin x cos x " as L C M}$

=>. (1 - cos^2 x) / (sin x cos x), color(crimson)(" identity " sin^2x + cos^2x = 1

$\implies {\sin}^{2} \frac{x}{\sin x \cos x}$

$\implies \sin \frac{x}{\cos} x = \tan x = R H S$