$\displaystyle \frac{sec x + 1}{tan x} = \frac{tan x}{sec x -1} $
x there is θ. I don't know how to put the θ.
Thank you.!
Substitute $\displaystyle \tan x = \frac{\sin x}{\cos x}$ and $\displaystyle \sec x = \frac{1}{\cos x}$ and simplify:
LHS $\displaystyle = \frac{1 + \cos x}{\sin x}$
RHS $\displaystyle = \frac{\sin x}{1 - \cos x} = \frac{\sin x}{(1 - \cos x)} \cdot \frac{(1 + \cos x)}{(1 + \cos x)} = \frac{\sin x (1 + \cos x)}{(1 - \cos^2 x)} $ and it should be clear how to continue and end up with the LHS.
Pardon me, but what is LHS and RHS?
OK, so I get the process of ending up with the LHS you were saying. But in our class, we need to retain the original term of the left or right side. So I need to show that $\displaystyle \frac{sec x + 1}{tan x} = \frac{sec x + 1}{tan x}
$
So from the LHS, can you show me how to make the LHS into $\displaystyle \frac{sec x + 1}{tan x}$