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solving from one side,

tan2x-1/sec2x = tanx-cotx/tanx+cotx 

 Feb 9, 2021
 #1
avatar+113775 
+1

Should there be any brackets here?

Please make it clear what the numerator and denominators are.

 Feb 10, 2021
 #2
avatar+113775 
+1

Please post your repaired question here. 

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

 

-----------------------

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.

 Feb 10, 2021
 #3
avatar+42 
+1

Ok first off apologies, I didn't mean to sound or be rude, I just started this website today and I didn't know I could reply to what you said so I redid the problem and posted it again. 

I told you what the denominator and numerator were. Should I take a picture of the problem? Would that be better? In the problem there are no brackets so I'm not sure were the brackets would go....

Country03  Feb 10, 2021
edited by Country03  Feb 10, 2021
 #4
avatar+113775 
+1

Apology accepted.  Welcome to Web2.0 calc forum.   laugh

 

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.  :)

 Feb 10, 2021
edited by Melody  Feb 10, 2021
edited by Melody  Feb 10, 2021
 #5
avatar+42 
0

Thank you :) Sorry again 

It's B

Country03  Feb 10, 2021
 #6
avatar+113775 
0

It is a tough one.   frown

Melody  Feb 10, 2021
 #7
avatar+42 
0

Yea ūüė¨

Country03  Feb 10, 2021
 #8
avatar+113775 
+1

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

 Feb 10, 2021
 #9
avatar+32311 
+4

I think the denominator on the LHS is sec2(x), not sec(2x), so we have:

 

Alan  Feb 10, 2021
 #10
avatar+113775 
+1

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)

 Feb 10, 2021
edited by Melody  Feb 10, 2021
edited by Melody  Feb 10, 2021
edited by Melody  Feb 10, 2021

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