tan left( 90^o - cot^{-1} frac{1}{3} right) = | vedclass

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(C) We know that $	an(90^o - 	heta) = \cot(	heta)$.Substituting $	heta = \cot^{-1} \frac{1}{3}$,we get:$	an \left( 90^o - \cot^{-1} \frac{1}{3} \

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(C) We know that $	an(90^o - 	heta) = \cot(	heta)$.Substituting $	heta = \cot^{-1} \frac{1}{3}$,we get:$	an \left( 90^o - \cot^{-1} \frac{1}{3} ight) = \cot \left( \cot^{-1} \frac{1}{3} ight)$.Since $\cot(\cot^{-1} x) = x$ for all $x \in \mathbb{R}$,we have:$\cot \left( \cot^{-1} \frac{1}{3} ight) = \frac{1}{3}$.Thus,the correct option is $C$.

$\tan \left( 90^o - \cot^{-1} \frac{1}{3} \right) = $

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