Let me rewrite the problems because I see where they could be confusing. I will put the work I have done too, but I have tried multiple routes for each problem.
Prove that the following identity is true: 1/(cscx-sinx) = tanx*secx
Work:
I decided to work with the left side of my equation which I am not even sure is the best way.
Try the same technique for the next problem
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Whenever you have trouble with trig-identity - change "everything" to sin(x) and cos(x). So the first problem becomes:
\(\displaystyle \frac{1}{csc(x)-sin(x)}\)
\(\displaystyle = \ \frac{1}{\frac{1}{sin(x)}-sin(x)}\)
\(\displaystyle = \ \frac{1}{\frac{1 \ - \ \sin^2(x)}{sin(x)}}\)
\(\displaystyle = \ \frac{sin(x)}{1 \ - \ \sin^2(x)}\)
\(\displaystyle = \ \frac{sin(x)}{cos^2(x)}\)
\(\displaystyle = \ \frac{sin(x)}{cos(x)} \ * \ \frac{1}{cos(x)}\)
\(\displaystyle = \ tan(x) \ * \ sec(x)\)
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First I changed cscx to secx/tanx and changed sinx into cosx/cotx.
That gave me this equation:
(tanx-cotx)/(secx-cosx)
Then I changed cosx into sinx/tanx.
That gave me:
(tan^2x-cotx*tanx)/(secx-sinx)
I changed tan^2x to sec^2x-1 which allowed me to cancel out the secx in the denominator and one of the secx in the numerator.
So it left me with:
(secx-1-cotx*tanx)/-sinx
I know this is all getting me no where and really don't know what else to do.
Prove that the following identity is true: (cos^4x-sin^4x)/cos^2x = 2-sec^2x
Work:
I started to work out the left side because it was the most complex side, but gave me more to work with and break down.
The first thing I did was factor my numerator on the left side which gave me:
(cos^2x+sin^2x)*(cos^2x-sin^2x)/cos^2x.
I don't really know what to do after that.
