How do you calculate $tan ({cot}^{- 1} (9))$ $tan(cot^-1 (9))$ ?

Question

2088 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-22T12:34:53

$tan ({cot}^{- 1} (9)) = \frac{1}{9}$ $tan(cot^(-1)(9))=1/9$

Let $cot θ = 9$ $cottheta=9$

Hence $θ = {cot}^{- 1} (9)$ $theta=cot^(-1)(9)$

and $tan ({cot}^{- 1} (9)) = tan θ = \frac{1}{cot θ} = \frac{1}{9}$ $tan(cot^(-1)(9))=tantheta=1/cottheta=1/9$

How do you calculate tan(cot−1(9))tan(cot^-1 (9))?