verifying identities: cotx/cscx - cscx/cotx = -tanxsinx

[jacqueline]

cotx/cscx - cscx/cotx = -tanxsinx

This is what I have so far, but I feel like I've done something wrong...
cotx/cscx - cscx/cotx
= cosx/sinx/1/sinx - 1/sinx/cosx/sinx
= cosx - 1/cosx
= cosx - (1/cosx) sinx/sinx
= cosx - sinx/cosxsinx
= cosx - sinx/cosx (1/sinx)
= cosx - tanx (1/sinx)

 

[morson]

\(\displaystyle \frac{cotx}{cscx}\ = cosx\)

\(\displaystyle \frac{cscx}{cotx}\ = \frac{1}{cosx}\\)

You've established these results already.

So, the left hand side is: \(\displaystyle cosx - \frac{1}{cosx}\ = \frac{(cos^2x - 1)}{cosx}\ = -\frac{(1 - cos^2x)}{cosx}\ = -\frac{sin^2x}{cosx}\\), provided \(\displaystyle cotx, cscx, cosx \not=\ 0\)

 

[soroban]

Re: verifying identities

Hello, jacqueline!

\(\displaystyle \L\frac{\cot x}{\csc x}\,-\,\frac{\csc x}{\cot x}\: =\: -\tan x\cdot\sin x\)

Combine fractions: \(\displaystyle \L\:\frac{\cot^2x\,-\,\csc^2x}{\csc x\cot x}\)


Identity: \(\displaystyle \,\csc^2\theta\:=\:\cot^2\theta\,+\,1\;\;\Rightarrow\;\;\cot^2\theta\,-\,\csc^2\theta\:=\:-1\)

. . And we have: \(\displaystyle \L\:\frac{-1}{\csc x\cot x}\:=\:-\frac{1}{\cot x}\cdot\frac{1}{\csc x}\)


Since \(\displaystyle \frac{1}{\csc\theta}\,=\,\sin x\) and \(\displaystyle \frac{1}{\cot x} \,=\,\tan x\)

. .we have: \(\displaystyle \L\:-\tan x\cdot\sin x\)