# How do you prove csc B/cos B - cosB/sin B = tan B?

Nov 26, 2015

It is explained below

#### Explanation:

$= \frac{\csc B \sin B - {\cos}^{2} B}{\sin B \cos B}$

=(1-cos^2 B)/(sin B cos B)#

= $\frac{{\sin}^{2} B}{\sin B \cos B}$
= $\frac{\sin B}{\cos} B = \tan B$

Nov 26, 2015

$\csc \frac{B}{\cos} B - \cos \frac{B}{\sin} B = \tan B$

LS:
$\csc \frac{B}{\cos} B - \cos \frac{B}{\sin} B$

$= \frac{\frac{1}{\sin} B}{\cos} B - \cos \frac{B}{\sin} B$

$= \left(\frac{1}{\sin} B \div \cos B\right) - \cos \frac{B}{\sin} B$

$= \left(\frac{1}{\sin} B \cdot \frac{1}{\cos} B\right) - \cos \frac{B}{\sin} B$

$= \frac{1}{\sin B \cos B} - \cos \frac{B}{\sin} B$

$= \frac{1 - \cos B \left(\cos B\right)}{\sin B \cos B}$

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

$= {\sin}^{2} \frac{B}{\sin B \cos B}$

$= \frac{\sin B \sin B}{\sin B \cos B}$

$= \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\sin B}}} \sin B}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\sin B}}} \cos B}$

$= \tan B$

$\therefore$, LS = RS.