$\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.!

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- Oct 20th 2009, 12:57 AMreiwardProving identitiees
$\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.! - Oct 20th 2009, 02:25 AMmr fantastic
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. - Oct 20th 2009, 02:35 AMreiward
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}

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So from the LHS, can you show me how to make the LHS into $\displaystyle \frac{sec x + 1}{tan x}$ - Oct 20th 2009, 02:45 AMmr fantastic