• # question_answer Direction: For the following questions, choose the correct answer from the codes [a], [b], [c] and [d] defined as follows. Let us define the function as ${{\cos }^{-1}}\,(\cos \theta )=\theta$and$2{{\tan }^{-1}}x=\frac{2x}{1-{{x}^{2}}}$. Statement I If $\sin \,[2\,{{\cos }^{-1}}\{\cot \,(2\,{{\tan }^{-1}}x)\}]=0,$ then $x=\pm 1,\,\,\pm \,(1\pm \sqrt{2})$ Statement II $\cot \,\,(2\,{{\tan }^{-1}}x)=\frac{1-{{x}^{2}}}{2x}$ A)  Statement I is true. Statement II is also true and Statement II is the correct explanation of Statement I. B)  Statement I is true. Statement II is also true and Statement II is not the correct explanation of Statement I. C)  Statement I is true. Statement II is false. D)  Statement I is false. Statement II is true.

$\sin \,\theta =0\,\,\,\Rightarrow \,\,\,\theta n\pi$ $\Rightarrow$            $2\,{{\cos }^{-1}}[\cot (2\,{{\tan }^{-1}}x)]\,=n\pi$ $\Rightarrow$            ${{\cos }^{-1}}\,[\cot \,(2\,{{\tan }^{-1}}x)]=\frac{n\pi }{2}$ $\Rightarrow$            $\cot \,(2\,{{\tan }^{-1}}x)=\cos \,\frac{n\pi }{2}$ $\Rightarrow$            $\frac{1}{\tan \,\left( {{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}} \right)}=\cos \,\frac{n\pi }{2}$ $\Rightarrow$            $\frac{1-{{x}^{2}}}{2x}=\cos \,\frac{n\pi }{2}$ Now, $\cos \,\frac{n\pi }{2}=0,$ if $n=1,\,\,3,\,5,...$ $\therefore$    $\frac{1-{{x}^{2}}}{2x}=0\,\,\Rightarrow \,\,x=\pm 1$ Now, $\cos \,\frac{n\pi }{2}=1,$ if $n=0,4,...$ $\therefore$    $\frac{1-{{x}^{2}}}{2x}=1\Rightarrow \,x=-1\,\pm \,\sqrt{2}$ Now, $\cos \,\frac{n\pi }{2}=-1,$ if $n=2,\,\,6,...$ $\therefore$    $\frac{1-{{x}^{2}}}{2x}=-1\,\,\,\Rightarrow \,\,x=1\pm \sqrt{2}$ Hence, $x=\pm 1,\,\,\pm 1\pm \sqrt{2}$