If {cot ^{ - 1}}frac{n}{pi } frac{pi }{6},, | vedclass

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

Question

(C) Given the inequality: ${\cot ^{ - 1}}\frac{n}{\pi } > \frac{\pi }{6}$.Since the function $f(x) = {\cot ^{ - 1}}x$ is a strictly decreasing functio

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given the inequality: ${\cot ^{ - 1}}\frac{n}{\pi } > \frac{\pi }{6}$.Since the function $f(x) = {\cot ^{ - 1}}x$ is a strictly decreasing function,the inequality sign reverses when we take the cotangent of both sides:$\frac{n}{\pi } We know that $\cot \left( \frac{\pi }{6} \right) = \sqrt{3}$.So,$\frac{n}{\pi } Multiplying both sides by $\pi$:$n Using the approximation $\pi \approx 3.14159$ and $\sqrt{3} \approx 1.732$:$n Since $n \in N$ (natural numbers),the maximum integer value of $n$ satisfying $n < 5.44$ is $5$.

If ${\cot ^{ - 1}}\frac{n}{\pi } > \frac{\pi }{6},\,\,n \in N$,then the maximum value of $n$ is

Similar Questions

If $\tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7)$,then $x$ is equal to

$\sin [\cot ^{ - 1}(\cos \tan ^{ - 1}x)] =$

$\begin{aligned} & \text{If } \cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\} \\ & b>a, \text{ then } x= \end{aligned}$

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{70} + \tan^{-1} \frac{1}{99} = $

If $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2},$ then the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module