# How do you show the secx + 1 + (1 - tan^2x)/(secx -1) = cosx/(1 - cosx)?

Apr 7, 2018

We have:

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

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

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

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

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

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

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

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

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

$L H S = R H S$

Hopefully this helps!

Apr 7, 2018

To prove the identity, we'll need these identities:

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

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

I'll start with the left side of the equation and manipulate it until it equals the right side:

$L H S = \sec x + 1 + \frac{1 - {\tan}^{2} x}{\sec x - 1}$

$\textcolor{w h i t e}{L H S} = \frac{\left(\sec x + 1\right) \left(\sec x - 1\right)}{\sec x - 1} + \frac{1 - {\tan}^{2} x}{\sec x - 1}$

$\textcolor{w h i t e}{L H S} = \frac{\textcolor{b l u e}{{\sec}^{2} x} - 1}{\sec x - 1} + \frac{1 - {\tan}^{2} x}{\sec x - 1}$

$\textcolor{w h i t e}{L H S} = \frac{\textcolor{b l u e}{{\tan}^{2} x + 1} - 1}{\sec x - 1} + \frac{1 - {\tan}^{2} x}{\sec x - 1}$

$\textcolor{w h i t e}{L H S} = \frac{{\tan}^{2} x \textcolor{red}{\cancel{\textcolor{b l a c k}{\textcolor{b l a c k}{+} 1 - 1}}}}{\sec x - 1} + \frac{1 - {\tan}^{2} x}{\sec x - 1}$

$\textcolor{w h i t e}{L H S} = \frac{{\tan}^{2} x}{\sec x - 1} + \frac{1 - {\tan}^{2} x}{\sec x - 1}$

$\textcolor{w h i t e}{L H S} = \frac{{\tan}^{2} x + 1 - {\tan}^{2} x}{\sec x - 1}$

$\textcolor{w h i t e}{L H S} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{\tan}^{2} x}}} + 1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\textcolor{b l a c k}{-} {\tan}^{2} x}}}}{\sec x - 1}$

$\textcolor{w h i t e}{L H S} = \frac{1}{\textcolor{b l u e}{\sec} x - 1}$

$\textcolor{w h i t e}{L H S} = \frac{1}{\textcolor{b l u e}{\frac{1}{\cos} x} - 1}$

$\textcolor{w h i t e}{L H S} = \frac{1}{\left(\frac{1}{\cos} x - 1\right)} \textcolor{red}{\cdot \cos \frac{x}{\cos} x}$

$\textcolor{w h i t e}{L H S} = \cos \frac{x}{1 - \cos x}$

$\textcolor{w h i t e}{L H S} = R H S$

That's the proof. Hope this helped!