Please help

Should there be any brackets here?

Please make it clear what the numerator and denominators are.

Please post your repaired question here. 

Don't just ignore me and repeat the question elsewhere, that is rude.

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This is what you said in your new question:

I have to vertify the identity (there are no brackets in this problem) 

tan2x-1 (numerator) / sec2x (denominator) = tanx - cotx (numerator) / tanx + cotx (denominator) 

I still do not know what your question is, 

for example, is the numerator  tan2x-1  or     is it   -1   with the 2x seperate?   

That is what brackets are for.  Brackets would make the meaning clear.

If you have copied this from something written with fraction lines then the fraction line is used instead of brackets, if you take away the fraction line you must introduce brackets.

Apology accepted.  Welcome to Web2.0 calc forum.   

No you didn't tell me

You said:

tan2x-1 (numerator) / sec2x (denominator) = tanx - cotx (numerator) / tanx + cotx (denominator) 

Maybe this means

A       tan2x- (1  / sec2x) = tanx -   [cotx / (tanx + cotx )]

or

B       (tan2x-1 )  / sec2x  = (tanx - cotx ) / (tanx + cotx)

Is it A or is it B ?

It is important that you learn to go between fractions and brackets.

If you are still confused then take a pic but I would like you to tell me if it is A or B anyway.  :)

I think I have proven that they are not equal.  Tell me if you see any errors in my woking.

Maybe my interpretation of your question was still not what you intended to ask?

\(\frac{tan2x-1}{sec2x} =\frac{ tanx-cotx}{tanx+cotx }\\~\\ RHS=\left[\frac{sinx}{cosx}-\frac{cosx}{sinx} \right] \div \left[\frac{sinx}{cosx}+\frac{cosx}{sinx} \right]\\ RHS=\left[\frac{sin^2x-cos^2x}{cosxsinx} \right] \div \left[\frac{sin^2x+cos^2x}{cosxsinx}\right]\\ RHS=\left[\frac{sin^2x-cos^2x}{cosxsinx} \right] \div \left[\frac{1}{cosxsinx}\right]\\ RHS=sin^2x-cos^2x\\ RHS=-(cos^2x-sin^2x)\\ RHS=-cos(2x)\\ RHS=\frac{-1}{sec(2x)}\\ RHS \ne LHS\)

LaTex:

\frac{tan2x-1}{sec2x} =\frac{ tanx-cotx}{tanx+cotx }\\~\\
RHS=\left[\frac{sinx}{cosx}-\frac{cosx}{sinx}  \right] \div \left[\frac{sinx}{cosx}+\frac{cosx}{sinx}  \right]\\
RHS=\left[\frac{sin^2x-cos^2x}{cosxsinx} \right] \div \left[\frac{sin^2x+cos^2x}{cosxsinx}\right]\\
RHS=\left[\frac{sin^2x-cos^2x}{cosxsinx} \right] \div \left[\frac{1}{cosxsinx}\right]\\
RHS=sin^2x-cos^2x\\
RHS=-(cos^2x-sin^2x)\\
RHS=-cos(2x)\\
RHS=\frac{-1}{sec(2x)}\\
RHS \ne LHS

ok Thanks Alan,

Sorry Country03, I see your original question did show this clearly, so it was my error.

but when I asked if  A or B was correct you said B, which was not what you wanted at all.

This is what you wanted.

[ (tanx)^2 -1 ]  / (secx)^2  = (tanx - cotx ) / (tanx + cotx)