How would you prove tanx/cscx = secx - 1/secx?

May 7, 2016

Replace all simple functions
$\tan \frac{x}{\csc} x = \tan x \sin x = {\sin}^{2} \frac{x}{\cos} x$
$\frac{1}{\cos} x - \frac{1}{\sec} x = \frac{1}{\cos} x - \cos x = {\sin}^{2} \frac{x}{\cos} x$

May 9, 2016

Just to clarify the answer of the previous contributor.

Explanation:

On the left side, we must apply the identities $\tan x = \sin \frac{x}{\cos} x$ and $\csc x = \frac{1}{\sin} x$

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

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

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

Applying the Pythagorean identity $1 - {\cos}^{2} x = {\sin}^{2} x$

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

Identity proved!!

Hopefully this helps!