Establish the Identity

[drummguy04]

(sinx/cscx-1)+(sinx/cscx+1)=2tan^2x

I have tried to prove the equation by starting on the left side and working to get rid of the Cosecants. Unfortunately it hasn't gone well.

 

[HallsofIvy]

First, you will need to clarify, is it (sinx/(cscx-1) )+(sinx/(cscx+1))=2tan^2x?

I woud be inclined to change everything to sine and cosine: sin(x)/(1/sin(x)- 1)+sin(x)/(1/sin(x)+1). Simplify by multiplying both numerator and denominator by sin(x): sin^2(x)/(1- sin(x))+ sin^2(x)/(1+ sin(x)). To combine the fractions we need to get a "common denominator" by multiplying numerator and denominator of the first fraction by 1+ sin(x) and of the second fraction by 1- sin(x):
$\displaystyle \frac{sin^2(x)(1+ sin(x))}{1- sin^2(x)}+ \frac{sin^2(x)(1- sin(x))}{1- sin^2(x)}$
The denominator is, of course, $\displaystyle cos^2(x)$ so that becomes $\displaystyle \frac{sin^2(x)+ sin^3(x)+ sin^2(x)- sin^3(x)}{cos^2(x)}= \frac{2sin^2(x)}{cos^2(x)}$.

 

[soroban]

Hello, drummguy04!

$\displaystyle \dfrac{\sin x}{\csc x-1}+\dfrac{\sin x}{\csc x+1}\:=\:2\tan^2\!x$

I would do it like this . . .

$\displaystyle \displaystyle\dfrac{\sin x}{\csc x - 1} + \dfrac{\sin x}{\csc x + 1} \;=\; \sin x\left(\frac{1}{\csc x - 1} + \frac{1}{\csc x + 1}\right)$

. . . . . . . . . . . . . . . . . $\displaystyle \displaystyle=\;\sin x\left(\frac{(\csc x+1) + (\csc x-1)}{(\csc x -1)(\csc x + 1)}\right)$

. . . . . . . . . . . . . . . . . $\displaystyle =\;\dfrac{\sin x\cdot2\csc x}{\csc^2\!x-1}$

. . . . . . . . . . . . . . . . . $\displaystyle =\;\dfrac{2}{\cot^2\!x}$

. . . . . . . . . . . . . . . . . $\displaystyle =\;2\tan^2\!x$

 

[drummguy04]

Thank you for your help.