tan identity

**jacs** · September 23rd 2010, 09:29 PM

I have this question
Express tan (90 + x) in terms of tan x

not really sure what to do with it, using the compound angles i have gotten this far

$\frac{tan90+tanx}{1-tan90tanx}$

= $\frac{\infty+tanx}{1-\infty tanx}$

not really sure how to proceed from there, or if that is an acceptable answer, or i am going about this in completely the wrong way

thanks

**TheCoffeeMachine** · September 23rd 2010, 10:20 PM

The identity $\tan(a+b) = \frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}$ is not valid if either $a$ or $b$ is equal to $\frac{\pi}{2}$ .

Why? Because if, for example, $a = \frac{\pi}{2}$ , then $\tan{a} = \tan\left(\frac{\pi}{2}}\right) = \tan\left(\frac{\pi}{4}+\frac{\pi}{4}\right)$ .

Using the same identity we would get $\tan{a} = \frac{\tan{\frac{\pi}{4}}+\tan{\frac{\pi}{4}}}{1-\tan{\frac{\pi}{4}}\tan{\frac{\pi}{4}}}$ .

But since $\tan{\frac{\pi}{4}} = 1$ , our denominator is zero. Therefore $\tan{\frac{\pi}{2}}$ is undefined.

Don't worry though. By symmetry, or otherwise, you can get your $\tan\left(\frac{\pi}{2}+x\right)[/Math] to [Math]-\cot{x}.$

**jacs** · September 25th 2010, 12:35 AM

I think i have figured out a way to do it without looking at the graphs, although i now see the transformation from tan x to tan (90 + x) = - cot x.

Is this a sound approach?

180 - (90 - x) = 90 + x
hence
tan (90 + x) = tan {180 - (90 - x)}
= -tan (90 - x) (lying in the second quadrant)
= - cot x (complementary angles)
= $-\frac{1}{tanx}$

thus expressing the original in terms of x

**TheCoffeeMachine** · September 25th 2010, 12:51 AM

Looks good to me.

More neater perhaps would be to write:

$\displaystyle \tan\left(\frac{\pi}{2}+x\right) = \frac{\sin(\frac{\pi}{2}+x)}{\cos(\frac{\pi}{2}+x) } [/Math] [Math] \displaystyle = \frac{\sin\frac{\pi}{2}\cos{x}+\cos\frac{\pi}{2}\s in{x}}{\cos\frac{\pi}{2}\cos{x}-\sin\frac{\pi}{2}\sin{x}} = \frac{(1)\cos{x}+(0)\sin{x}}{(0)\cos{x}-(1)\sin{x}} = -\frac{\cos{x}}{\sin{x}} = -\frac{1}{\tan{x}}.$

Looks hot too.

**jacs** · September 25th 2010, 12:56 AM

Hadn't even thought of rewriting it in terms of sin and cos, and now i see it, it is so glaringly obvious.

Thanks so much for your help with this problem, i now have 3 distinct ways of looking at it,which is three more than i had yesterday, lol

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