# Math Help - Trigonomtertic Identity help

1. ## Trigonomtertic Identity help

I need help proving the following trig identity: ( x stands for theta)
csc x -1 / cot x = cot x / csc x + 1

2. Originally Posted by mathmike
I need help solving the following trig identity: ( x stands for theta)
csc x -1 / cot x = cot x / csc x + 1
solving or proving? and do you mean $\displaystyle \frac {\csc x - 1}{\cot x} = \frac {\cot x}{\csc x + 1}$ ?

Hint: change everything to sines and cosines, and show you can simplify one side to get the other.

3. Hello, mathmike!

Are familiar with this identity and its variations?

. $\csc^2\!\theta \:=\:\cot^2\!\theta + 1 \quad\Rightarrow\quad \csc^2\!\theta -1\:=\:\cot^2\!\theta \quad\Rightarrow \quad \csc^2\!\theta - \cot^2\!\theta \:=\:1$

Prove: . $\dfrac{\csc x -1}{\cot x} \:=\: \dfrac{\cot x}{\csc x + 1}$

The right side is: . $\dfrac{\cot x}{\csc x + 1}$

Multiply by $\frac{\csc x - 1}{\csc x - 1}\!:\;\;\dfrac{\cot x}{\csc x + 1}\cdot\dfrac{\csc x - 1}{\csc x - 1} \;=\; \dfrac{\cot x(\csc x - 1)}{\csc^2\!x - 1}$

. . . . . . . . . . . . $=\;\dfrac{\cot x(\csc x-1)}{\cot^2\!x} \;=\;\dfrac{\csc x - 1}{\cot x}$