The value of $sin \left[cot^{-1}\begin{Bmatrix}cos(tan^{-1}1)\end{Bmatrix}\right]$, is
Answer & explanation
Correct answer: option 2
$sin \left[cot^{-1}\begin{Bmatrix}cos(tan^{-1}1)\end{Bmatrix}\right]$
$=sin\begin{Bmatrix}cot^{-1}\left(cos\frac{\pi}{4}\right)\end{Bmatrix}$ $\left[∵tan^{-1}1=\frac{\pi}{4}\right]$
$=sin\left(cot^{-1}\frac{1}{\sqrt{2}}\right)$
$=sin\left(sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}\right)=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{\frac{2}{3}}$ $\left[∵cot^{-1}\frac{1}{\sqrt{2}}=sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}\right]$