## Pawanyadav one year ago -1 -1 2cos{cot(2tan x)}=0, Then possible values of x can be: (A) √2+1 (B)2+√3 (C)sign(Π) (D)1-√2

1. anonymous

cos (x)=0 => x= pi/2

2. anonymous

@Pawanyadav Is that supposed to say $2\cos^{-1}\left(\cot\left(2\tan^{-1}x\right)\right)~?$

3. anonymous

$\cot2t=\frac{\cos2t}{\sin2t}=\frac{\cos^2t-\sin^2t}{2\sin t\cos t}$ Replacing $$t=\tan^{-1}x$$, you have $$\tan t=\dfrac{x}{1}$$. |dw:1433859845076:dw| This suggests that $$\sin t=\dfrac{x}{\sqrt{1+x^2}}$$ and $$\cos t=\dfrac{1}{\sqrt{1+x^2}}$$. \begin{align*}2\cos^{-1}\left(\cot\left(2\tan^{-1}x\right)\right)&=2\cos^{-1}\left(\frac{\cos^2t-\sin^2t}{2\sin t\cos t}\right)\\\\ &=2\cos^{-1}\left(\frac{1}{2}\cot t-\frac{1}{2} \tan t\right)\end{align*} You started with $$2\cos^{-1}(\cdots)=0$$, which means $$\cdots=\cos0=1$$, so you have the equation $\frac{1}{2}\cot t-\frac{1}{2} \tan t=1$ Multiplying both sides by $$\tan t$$ (or $$\cot t$$) will give you a quadratic: \begin{align*}\frac{1}{2}-\frac{1}{2} \tan^2 t&=\tan t\\\\ \tan^2t+2\tan t-1&=0\\\\ \tan t&=\frac{-2\pm\sqrt{8}}{2}\\\\ &=-1\pm\sqrt 2 \end{align*} Recall that $$\tan t=x$$, so we're done.