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

My work

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 $(-(3)^1/2 /4 ,1/2)$

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

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 $$y=\frac{-\cos t-\sqrt{3}\sin t}{-\sqrt{3}\cos{t}+\sin t}=-\frac{\cos\left(t-\frac{\pi}{3}\right)}{\sin\left(t-\frac{\pi}{3}\right)}=-\cot\left(t-\frac{\pi}{3}\right)$$

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    $\begingroup$ How does it help? Can you explain. $\endgroup$ – Archis Welankar Jun 15 '16 at 17:06
  • $\begingroup$ What was the idea in your mind that you should replace it with this? $\endgroup$ – Aakash Kumar Jun 15 '16 at 17:07

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