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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
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos (x)=0 => x= pi/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Pawanyadav Is that supposed to say \[2\cos^{1}\left(\cot\left(2\tan^{1}x\right)\right)~?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cot2t=\frac{\cos2t}{\sin2t}=\frac{\cos^2t\sin^2t}{2\sin t\cos t}\] 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: \[\begin{align*}\frac{1}{2}\frac{1}{2} \tan^2 t&=\tan t\\\\ \tan^2t+2\tan t1&=0\\\\ \tan t&=\frac{2\pm\sqrt{8}}{2}\\\\ &=1\pm\sqrt 2 \end{align*}\] Recall that \(\tan t=x\), so we're done.
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