# Simplification of this function

$f(x)= ((1- 4x^2)^{1/2} - 2(3)^{1/2}x)/((3-12x^2)^{1/2} + 2x)$ Find range when $x$ belongs to $(-(3)^1/2 /4 ,1/2)$

I have to find the range of this function , I have simplified this expression , I want to know that can I simplify it further more.

Original question was $\tan^{-1}( ((1- 4x^2)^{1/2} - 2(3)^{1/2}x)/((3-12x^2)^{1/2} + 2x))$ Find range in $(-(3)^1/2 /4 ,1/2)$

• I got the range as $[\frac{-8}{\sqrt{3}},\frac{1}{\sqrt{3}}]$ – Archis Welankar Jun 15 '16 at 16:58
• Can you simplify this expression further ? – Aakash Kumar Jun 15 '16 at 17:00
• No but do you know the correct answer so that I can verify my answer because giving wrong answer will be pointless;) – Archis Welankar Jun 15 '16 at 17:03
• I know answer but the question was bit different , it was tan inverse(f(x)) – Aakash Kumar Jun 15 '16 at 17:06
• Can you post the original equation – Archis Welankar Jun 15 '16 at 17:13

let $x=\frac{1}{2}\sin t$ ,$\,-\frac{\pi}{2}\le t \le\frac{\pi}{2}\,$ we have $$y=\frac{\cos t-\sqrt{3}\sin t}{\sqrt{3}\cos{t}+\sin t}=\frac{\cos\left(t+\frac{\pi}{3}\right)}{\sin\left(t+\frac{\pi}{3}\right)}=\cot\left(t+\frac{\pi}{3}\right)$$ let $x=\frac{1}{2}\sin t$ ,$\,\frac{\pi}{2}< t \le \pi\,$ we have $$y=\frac{-\cos t-\sqrt{3}\sin t}{-\sqrt{3}\cos{t}+\sin t}=-\frac{\cos\left(t-\frac{\pi}{3}\right)}{\sin\left(t-\frac{\pi}{3}\right)}=-\cot\left(t-\frac{\pi}{3}\right)$$