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Why wouldn't this work?

If we were asked to find the expression in terms of x for tan(arc cosx)

If I did
Let y = arccos x
cos y = x

1cot2y=1sin21 \frac{1}{cot^2 y} = {\frac{1}{sin^2} - 1}

tany=11x21 tan y = \sqrt {\frac{1}{\sqrt{1-x^2}} - 1}
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By using tan = 1/cot and cot^2 + 1 = cosex^2
Have I made an error or does this method not work?
Also would this easily be simplified or should I leave it as it is?

Thank you!
Original post by IgorYakov


1cot2y=1sin21 \frac{1}{cot^2 y} = {\frac{1}{sin^2} - 1}


Where did you get this formula from? It doesn't look right.
Original post by ghostwalker
Where did you get this formula from? It doesn't look right.


Indeed. It's actually cot2y=1sin2y1\cot^2y = \frac{1}{\sin^2y} -1
Original post by FireGarden
Indeed. It's actually cot2y=1sin2y1\cot^2y = \frac{1}{\sin^2y} -1


OK.

Now how did you derive your formula for tan y? There appears to be one too many square roots there. I don't think the inner one should be there.
Reply 4
Original post by ghostwalker
OK.

Now how did you derive your formula for tan y? There appears to be one too many square roots there. I don't think the inner one should be there.



Aah yeh that's supposed to be what the other poster said, that's where my mistake was, so would the actual answer be:

x2+1x2 \sqrt \frac{x^2 + 1}{x^2}

and then I can simplify that with just an x as the denominator?
(edited 9 years ago)
Original post by IgorYakov
Aah yeh that's supposed to be what the other poster said, that's where my mistake was, so would the actual answer be:

x2+1x2 \sqrt \frac{x^2 + 1}{x^2}

and then I can simplify that with just an x as the denominator?


Yes, you can then take the 1/x^2 out as 1/x.
Reply 6
Original post by ghostwalker
Yes, you can then take the 1/x^2 out as 1/x.



Thank you!

Wont let me rep :redface:

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