Question

show that tan(x)+sec(x)-1/tan(x)+sec(x)-1=1+sin(x)/cos(x)

Answer 1

We have reviewed and slightly altered the problem because it was incorrect as given:

Show that

$$\frac {\tan x + \sec x - 1}{\tan x + \sec x} - 1 = - \frac {\cos x}{1 + \sin x}$$

Solution.

Start with the left side. Decompose into summands:

$$\frac {\tan x + \sec x - 1}{\tan x + \sec x} - 1 = \frac {\tan x + \sec x}{\tan x + \sec x} - \frac {1}{\tan x + \sec x} - 1 = 1 - \frac {1}{\tan x + \sec x} - 1 = - \frac {1}{\tan x + \sec x}$$

We know that $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. Then transform the expression:

$$- \frac {1}{\tan x + \sec x} = - \frac {1}{\frac {\sin x}{\cos x} + \frac {1}{\cos x}} = - 1 / \left(\frac {\sin x + 1}{\cos x}\right) = - \frac {\cos x}{\sin x + 1}$$

So we have identity.