simplifying inverse and normal trig functions

**mer1988** · August 29th 2007, 09:20 PM

i really need to know how to simplify this equation for -infinity<t<infinity,
cos(arctan(t))...

thanks

**red_dog** · August 29th 2007, 10:43 PM

Let $\arctan t=\theta\Rightarrow \tan\theta =t,\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$
So we have to find $\cos\theta$ , knowing $\tan\theta$ .
We have $\displaystyle\cos\theta=\pm\frac{1}{\sqrt{1+\tan^2 \theta}}$
On the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ , $\cos\theta\geq 0$ , so we choose the sign $+$
Thus, $\cos(\arctan t)=\frac{1}{\sqrt{1+t^2}}$

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