Question

-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

Answer 1

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

Answer 2

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

Answer 3

\[\cot2t=\frac{\cos2t}{\sin2t}=\frac{\cos^2t-\sin^2t}{2\sin t\cos t}\]
Replacing \(t=\tan^{-1}x\), you have \(\tan t=\dfrac{x}{1}\).

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.

Answer 4

Answer are. A,C,D