Consider the statements:(I) If f(x) = sin lef | vedclass

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(A) $(I)$ We have $f(x) = \sin \left(\cot ^{-1} \left(\cos \left(	an ^{-1} xight)ight)ight)$.Let $	an ^{-1} x = 	heta$, then $	an 	heta = x

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Both $I$ and $II$ are false

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(A) $(I)$ We have $f(x) = \sin \left(\cot ^{-1} \left(\cos \left(	an ^{-1} xight)ight)ight)$.Let $	an ^{-1} x = 	heta$, then $	an 	heta = x$. Thus, $\cos 	heta = \frac{1}{\sqrt{1+x^2}}$.So, $f(x) = \sin \left(\cot ^{-1} \left(\frac{1}{\sqrt{1+x^2}}ight)ight)$.Let $\cot ^{-1} \left(\frac{1}{\sqrt{1+x^2}}ight) = \alpha$, then $\cot \alpha = \frac{1}{\sqrt{1+x^2}}$, which implies $	an \alpha = \sqrt{1+x^2}$.Then $\sin \alpha = \frac{	an \alpha}{\sqrt{1+	an^2 \alpha}} = \frac{\sqrt{1+x^2}}{\sqrt{1+1+x^2}} = \sqrt{\frac{1+x^2}{2+x^2}}$.Therefore, $f(0) = \sqrt{\frac{1+0}{2+0}} = \frac{1}{\sqrt{2}}$.Since $\frac{1}{\sqrt{2}} 
eq \frac{1}{2}$, Statement $I$ is false.$(II)$ We use the formula $4 	an ^{-1} \frac{1}{5} = 	an ^{-1} \frac{120}{119}$.Then $\sin \left(	an ^{-1} \frac{120}{119} - 	an ^{-1} \frac{1}{239}ight) = \sin \left(	an ^{-1} \left(\frac{\frac{120}{119} - \frac{1}{239}}{1 + \frac{120}{119} 	imes \frac{1}{239}}ight)ight) = \sin \left(	an ^{-1} \left(\frac{28680 - 119}{28441 + 120}ight)ight) = \sin \left(	an ^{-1} \frac{28561}{28561}ight) = \sin \left(	an ^{-1} 1ight) = \sin \left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$.Since $\frac{1}{\sqrt{2}} 
eq 1$, Statement $II$ is false.

Consider the statements:
$(I)$ If $f(x) = \sin \left(\cot ^{-1} \left(\cos \left(\tan ^{-1} x\right)\right)\right)$, then $f(0) = \frac{1}{2}$.
$(II)$ $\sin \left(4 \tan ^{-1} \frac{1}{5} - \tan ^{-1} \frac{1}{239}\right) = 1$.
Then the correct option among the following is:

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Consider the statements:$(I)$ If $f(x) = \sin \left(\cot ^{-1} \left(\cos \left(\tan ^{-1} x\right)\right)\right)$, then $f(0) = \frac{1}{2}$.$(II)$ $\sin \left(4 \tan ^{-1} \frac{1}{5} - \tan ^{-1} \frac{1}{239}\right) = 1$.Then the correct option among the following is: