Identify the correct order.

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

Question

(D) To determine the correct order,we evaluate the values of the inverse secant function:$1$. $\sec^{-1}(1) = 0$$2$. $\sec^{-1}(2) = \frac{\pi}{3} \ap

Vedclass · Accepted Answer

$sec^{-1}(1) < sec^{-1}(2) < sec^{-1}(-2) < sec^{-1}(-1)$

Vedclass · Answer

(D) To determine the correct order,we evaluate the values of the inverse secant function:$1$. $\sec^{-1}(1) = 0$$2$. $\sec^{-1}(2) = \frac{\pi}{3} \approx 1.047$$3$. $\sec^{-1}(-2) = \frac{2\pi}{3} \approx 2.094$$4$. $\sec^{-1}(-1) = \pi \approx 3.141$Comparing these values,we get $0 Thus,the correct order is $\sec^{-1}(1) < \sec^{-1}(2) < \sec^{-1}(-2) < \sec^{-1}(-1)$.

Similar Questions

${\left[ {\sin \left( {{{\tan }^{ - 1}}\frac{3}{4}} \right)} \right]^2} = $

If $\sin ^{-1}\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{4}-\frac{x^{4}}{8}+\ldots\right)=\frac{\pi}{6},$ where $|x| < 2,$ then the value of $x$ is

If $\frac{(x + 1)^2}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$,then $\sin^{-1}\left(\frac{A}{C}\right) = $

$\cot ^{-1}\left(2 \cos \left(2 \operatorname{cosec}^{-1}(\sqrt{2})\right)\right)=\ldots$

Find the principal value of $\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module