Question

plz give me this answer​

Answer 1

Answer:

The answer to this question is $\frac{sin{t}}{t}$

Step-by-step explanation:

$cot^{-1}{x} = \frac{\pi}{2} - tan^{-1}{x}\\So,\\F(s)= cot^{-1}{s}= \frac{\pi}{2} - tan^{-1}{s}\\\text{Also we know, }\\\mathcal{L} [ -t f(t) ] = \frac{d}{ds}F(s)\\So, \frac{d}{ds}F(s)= - [\frac{1}{(s^2+1})]\\ \mathcal{L}^{-1} [\frac{d}{ds}F(s)] = -t f(t)\\ t f(t) = sin(t)\\\\or,f(t) = \frac{sin{t}}{t}$