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

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(D) We know that $	an^{-1}(	an x) = x$ only if $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$.Since $2$ radians is approximately $114.6^\circ$, which lies

Vedclass · Accepted Answer

$-\cos 2$

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(D) We know that $	an^{-1}(	an x) = x$ only 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.Let $	heta = 	an^{-1}(	an 2)$. Then $	an 	heta = 	an 2$.Since $2$ is in the second quadrant, we use the identity $	an(x - \pi) = 	an x$. Thus, $	an 2 = 	an(2 - \pi)$.Since $2 - \pi \approx -1.14$, which is in the range $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have $	heta = 2 - \pi$.Therefore, $\cos(	an^{-1}(	an 2)) = \cos(2 - \pi)$.Using the identity $\cos(x - \pi) = \cos(\pi - x) = -\cos x$, we get $\cos(2 - \pi) = -\cos 2$.Thus, the correct value is $-\cos 2$.

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

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