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Verify the identity sec x csc2 x - csc2 x = sec x/1+cosx
- Thread starter smiley16
- Start date
Re: Trig
Welcome to the forums! It would be nice if you showed some of your work first so that we could help you locate any errors and hopefully you will be able to learn from them (if any).
For this question, try converting the left side into terms of sinx and cosx and combine it into one single fraction. See if you can continue on from there and if not, come back with any more questions! Oh, also, keep in mind of these two things:
\(\displaystyle sin^{2}x + cos^{2}x = 1 \quad\quad sin^{2}x = 1 - cos^{2}x \quad\quad cos^{2}x = 1 - sin^{2}x \quad\quad \mbox{All these are equivalent}\)
\(\displaystyle \mbox{Difference of squares} \: : \quad a^{2} - b^{2} = (a+b)(a-b)\)
Also, try using brackets when posting questions because they can be ambiguous sometimes. For this question, it's clear that you meant secx / (1 + cosx). But for example you can have 1/x+4x + 6 which you can interpret this as:
\(\displaystyle \frac{1}{x} + 4x + 6 \quad \mbox{or} \quad \frac{1}{x + 4x} + 6 \quad \mbox{or} \quad \frac{1}{x + 4x + 6}\)
Welcome to the forums! It would be nice if you showed some of your work first so that we could help you locate any errors and hopefully you will be able to learn from them (if any).
For this question, try converting the left side into terms of sinx and cosx and combine it into one single fraction. See if you can continue on from there and if not, come back with any more questions! Oh, also, keep in mind of these two things:
\(\displaystyle sin^{2}x + cos^{2}x = 1 \quad\quad sin^{2}x = 1 - cos^{2}x \quad\quad cos^{2}x = 1 - sin^{2}x \quad\quad \mbox{All these are equivalent}\)
\(\displaystyle \mbox{Difference of squares} \: : \quad a^{2} - b^{2} = (a+b)(a-b)\)
Also, try using brackets when posting questions because they can be ambiguous sometimes. For this question, it's clear that you meant secx / (1 + cosx). But for example you can have 1/x+4x + 6 which you can interpret this as:
\(\displaystyle \frac{1}{x} + 4x + 6 \quad \mbox{or} \quad \frac{1}{x + 4x} + 6 \quad \mbox{or} \quad \frac{1}{x + 4x + 6}\)
Re: Trig
I really appreciate your help, but am still very confused on what to do. We are asked to step-by-step make the left side of the equation equal to the right side. I really do not even know how to begin the problem. Thank you for your help!
I really appreciate your help, but am still very confused on what to do. We are asked to step-by-step make the left side of the equation equal to the right side. I really do not even know how to begin the problem. Thank you for your help!
Re: Trig
Like I said convert into terms of sinx and cosx:
\(\displaystyle secx \cdot csc^{2}x - csc^{2}x = \frac{1}{cosx \cdot sin^{2}x} - \frac{1}{sin^{2}x}\)
And combine into one fraction:
\(\displaystyle \frac{1}{cosx \cdot sin^{2}x} \quad - \quad \frac{1}{sin^{2}x} \cdot \frac{cosx}{cosx}\)
\(\displaystyle = \frac{1}{cosx \cdot sin^{2}x} - \frac{cosx}{cosx \cdot sin^{2}x}\)
etc. etc.
Now using the hints from the first post, can you proceed? If you are still stuck, show us your work so we can help you along the way.
Like I said convert into terms of sinx and cosx:
\(\displaystyle secx \cdot csc^{2}x - csc^{2}x = \frac{1}{cosx \cdot sin^{2}x} - \frac{1}{sin^{2}x}\)
And combine into one fraction:
\(\displaystyle \frac{1}{cosx \cdot sin^{2}x} \quad - \quad \frac{1}{sin^{2}x} \cdot \frac{cosx}{cosx}\)
\(\displaystyle = \frac{1}{cosx \cdot sin^{2}x} - \frac{cosx}{cosx \cdot sin^{2}x}\)
etc. etc.
Now using the hints from the first post, can you proceed? If you are still stuck, show us your work so we can help you along the way.
Re: Trig
Nope you will be back to where you started. As I said before, combine them into one fraction. You have the same denominator, so just simply put the numerators over the same denominator:
\(\displaystyle \frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}\)
Now, again, here are some hints that might prove useful:
\(\displaystyle sin^{2}x = 1 - cos^{2}x\)
\(\displaystyle \mbox{Difference of squares :} \:\: a^{2} - b^{2} = (a+b)(a-b)\)
Nope you will be back to where you started. As I said before, combine them into one fraction. You have the same denominator, so just simply put the numerators over the same denominator:
\(\displaystyle \frac{a}{c} - \frac{b}{c} = \frac{a - b}{c}\)
Now, again, here are some hints that might prove useful:
\(\displaystyle sin^{2}x = 1 - cos^{2}x\)
\(\displaystyle \mbox{Difference of squares :} \:\: a^{2} - b^{2} = (a+b)(a-b)\)
Re: Trig
Okay so I combined the terms like you said and got (1-cosx)/(cos x*sin(squared)x)...if that makes sense..i dont know how to do the squared sign but anyway so i made sin(squared)x equal to 1-cos(squared)x and then had (1-cosx)/cosx*(1-cos(squared)x)...so i cancelled the 1-cosx and now have 1/cosx*(1-cosx) and i need that to somehow equal secx/1+cosx
Okay so I combined the terms like you said and got (1-cosx)/(cos x*sin(squared)x)...if that makes sense..i dont know how to do the squared sign but anyway so i made sin(squared)x equal to 1-cos(squared)x and then had (1-cosx)/cosx*(1-cos(squared)x)...so i cancelled the 1-cosx and now have 1/cosx*(1-cosx) and i need that to somehow equal secx/1+cosx
Re: Trig
Ah ok much clearer.
\(\displaystyle \frac{1-cosx}{cosx(1-cos^{2}x)}\)
You are right until here. Ignoring the cosx in the denominator for the moment:
\(\displaystyle \frac{1-cosx}{1-cos^{2}x} \neq \frac{1}{1-cosx}\)
If you tried multiplying back, you wouldn't get what you started with: \(\displaystyle (1-cosx)(1-cosx) = 1 - 2cosx + cos^{2}x\)
To further generalize:
\(\displaystyle \frac{a-b}{a^{2}-b^{2}} \neq \frac{1}{a-b}\)
You do not just cancel term by term.
Back to your question, use that difference of squares formula on 1 - cos[sup:3jipd8qo]2[/sup:3jipd8qo]x in the denominator and see what you get.
Ah ok much clearer.
\(\displaystyle \frac{1-cosx}{cosx(1-cos^{2}x)}\)
You are right until here. Ignoring the cosx in the denominator for the moment:
\(\displaystyle \frac{1-cosx}{1-cos^{2}x} \neq \frac{1}{1-cosx}\)
If you tried multiplying back, you wouldn't get what you started with: \(\displaystyle (1-cosx)(1-cosx) = 1 - 2cosx + cos^{2}x\)
To further generalize:
\(\displaystyle \frac{a-b}{a^{2}-b^{2}} \neq \frac{1}{a-b}\)
You do not just cancel term by term.
Back to your question, use that difference of squares formula on 1 - cos[sup:3jipd8qo]2[/sup:3jipd8qo]x in the denominator and see what you get.