Proof help! 1/cscx - 1/cotx = cscx + cotx

[Sid]

Prove:

1/cscx -1/cotx = cscx+cotx

I am at a lost here and converted the LHS to:

1/ 1/sinx - 1 1/cosx/sinx = sinx - sinx/cosx

find LCD and I get

(sinx)(cosx)/cosx - sinx/cosx

Now I am lost and not sure how I can get to cscx+cotx! Even from a negitive to positive.

Thanks
Sid

 

[soroban]

Re: Proof help!

Hello, Sid!

No wonder you can't get it . . . It is not an identity!

\(\displaystyle \frac{1}{\csc x}\,-\,\frac{1}{\cot x}\;=\;\csc x\,+\,\cot x\)

Let \(\displaystyle x\,=\,\frac{\pi}{4}:\L\;\;\frac{1}{\csc\frac{\pi}{4}}\,-\,\frac{1}{\cot\frac{\pi}{4}} \;= \;\csc\frac{\pi}{4}\,+\,\cot\frac{\pi}{4}\)

and we get: \(\displaystyle \L\;\frac{1}{\sqrt{2}}\,-\,\frac{1}{1}\;=\;\sqrt{2}\,+\,1\;\;\Rightarrow\;\;\frac{1\,-\,\sqrt{2}}{\sqrt{2}}\;=\;\sqrt{2}\,+\,1\)

The left side is negative; the right side is positive . . .

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Perhaps the problem was: \(\displaystyle \L\,\frac{1}{\csc x\,-\,\cot x}\;=\;\csc x + \cot x\)


On the left side, multiply top and bottom by \(\displaystyle (\csc x\,+\,\cot x):\)

\(\displaystyle \L\;\;\frac{1}{\csc x\,-\,\cot x}\cdot\frac{\csc x\,+\,\cot x}{\csc x\,+\,\cot x} \;= \;\frac{\csc x\,+\,\cot x}{\csc^2x\,-\,\cot^2x}\;=\;\frac{\csc x\,+\,\cot x}{1}\;\)\(\displaystyle =\;\csc x\,+\,\cot x\)

 

[Sid]

You are correct. I messed up by writting it out as if I was multiplying csc and cot when it was subtraction. I see my error and many thanks!

Sid