The value of cos (tan ^{ - 1}(tan 2)) is | vedclass

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(C) We know that the property of inverse trigonometric functions states that $	an^{-1}(	an x) = x$ if $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$.Since

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$\cos 2$

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(C) We know that the property of inverse trigonometric functions states that $	an^{-1}(	an x) = x$ if $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$.Since $2$ radians is approximately $114.6^\circ$,which lies outside the principal value branch $(-\frac{\pi}{2}, \frac{\pi}{2}) \approx (-1.57, 1.57)$,we must adjust the expression.However,the expression given is $\cos(	an^{-1}(	an 2))$.Let $	heta = 	an^{-1}(	an 2)$. Then $	an 	heta = 	an 2$.This implies $	heta = 2 - \pi$ (since $2$ is in the second quadrant,$	an 2 = 	an(2 - \pi)$ and $2 - \pi \approx -1.14$,which is in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$).Thus,$\cos(	an^{-1}(	an 2)) = \cos(2 - \pi)$.Using the identity $\cos(	heta - \pi) = \cos(\pi - 	heta) = -\cos 	heta$,we get $\cos(2 - \pi) = \cos(2)$.Therefore,the value is $\cos 2$.

The value of $\cos (\tan ^{ - 1}(\tan 2))$ is

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