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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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) = $

Similar Questions

Find the principal value of $cosec^{-1}(2)$.

If $\sin \left(\cot ^{-1}(x+1)\right)=\cos \left(\tan ^{-1} x\right)$,then considering positive square roots,$x$ has the value $\qquad$

Evaluate: $\tan ^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right)$

The sum of all the solution$(s)$ of the equation $\sin^{-1} 2x = \cos^{-1} x$ is

$1 + \cot^2(\sin^{-1}x) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module