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If x = sin 1 , y = cos 1 , z = tan 1 then

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x < y<z

z<x<y

y<z<x

y < x< z

answer is B.

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Detailed Solution

$1^{c} = 57^{0} \Rightarrow \sin 1 = \sin 57^{0}$ , $\cos 1 = \cos 57^{0}, \tan 1 = \tan 57^{0}$

$\cos 1 < \sin 1 < \tan 1 \Rightarrow y < x < z .$

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