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lkjs · Nov 13, 2009, 06:24 AM

ay I came this question:

expand and simplify:

tan( pi/2 - y)

I used the compound angle formula for tan(x-y), but don't know what do when I have tan( pi/2), since I take it to as being non-defined since its an asymptote, but the answer for the question comes as cot (y), can someone please explain and show the necessary steps taken to solve the equation. Thanks a lot.

P.S: I realise there is an exact question in a previous thread, but the reply wasn't very forthcoming. Thank you.

Unknown008 · Nov 13, 2009, 06:55 AM

I don't know if that is allowed, but I came up with this:

$tan(\frac{\pi}{2} - y) = \frac{tan\frac{\pi}{2} - tany}{1+tan\frac{\pi}{2}tany}$

$= \frac{\frac{sin \pi/2} {cos \pi/2 } - tany}{1+\frac{sin\pi/2}{cos\pi/2}{tany}}$

$= \frac{\frac{1} {0 } - tany}{1+\frac{1}{0}{tany}}$

$= \frac{\frac{1} {0 }}{1+\frac{1}{0}{tany}} - \frac{tany}{1+\frac{1}{0}{tany}}$

$= \frac{\frac{1} {0 }}{1+\frac{1}{0}{tany}} - \cancel{\frac{tany}{1+\frac{1}{0}{tany}}}$

$= \frac{\frac{1} {0 }}{\frac{0+tany}{0}}$

$= \frac{1}{ tany}$

I say again, that I don't know if that's the way to do it, but I came up with this... :o

s_cianci · Nov 13, 2009, 07:12 AM

Take it one step further ; 1/(tan y) = cot y.

ArcSine · Nov 13, 2009, 07:15 AM

Here's an approach relying on two of the 'co-function' identities...

$tan \left( \frac{\pi}{2} \ - \ y \right) \ = \ \frac{sin\left(\frac{\pi}{2} \ - \ y ) }{cos\left(\frac{\pi}{2} \ - \ y) } \ = \ \frac{\text{cos}(y)}{\text{sin}(y)} \ = \ \text{cot}(y)$

That second step is where the two co-function IDs kick in.

lkjs · Nov 13, 2009, 11:57 PM

Hey thanks for your replies, so the first solution is actually legitimate?

galactus · Nov 14, 2009, 05:30 AM

Originally Posted by Unknown008

I don't know if that is allowed, but I came up with this:

$tan(\frac{\pi}{2} - y) = \frac{tan\frac{\pi}{2} - tany}{1+tan\frac{\pi}{2}tany}$

$= \frac{\frac{sin \pi/2} {cos \pi/2 } - tany}{1+\frac{sin\pi/2}{cos\pi/2}{tany}}$

$= \frac{\frac{1} {0 } - tany}{1+\frac{1}{0}{tany}}$

$= \frac{\frac{1} {0 }}{1+\frac{1}{0}{tany}} - \frac{tany}{1+\frac{1}{0}{tany}}$

$= \frac{\frac{1} {0 }}{1+\frac{1}{0}{tany}} - \cancel{\frac{tany}{1+\frac{1}{0}{tany}}}$

$= \frac{\frac{1} {0 }}{\frac{0+tany}{0}}$

$= \frac{1}{ tany}$

I say again, that I don't know if that's the way to do it, but I came up with this... :o

Good use of LaTex. I like the cancel you used. You learned me something. I always wondered how to do that on this site. I tried a greenie, but it would not let me.

Perito · Nov 14, 2009, 05:30 AM

Yes. Both solutions are legitimate.

Unknown008 · Nov 14, 2009, 07:18 AM

Phew, I wasn't sure f we could 'play' with the zero like an ordinary number :o