Let x in (0, 1). The set of all x such that s | vedclass

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

Question

(B) Given the inequality $\sin^{-1} x > \cos^{-1} x$ for $x \in (0, 1)$.We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.Substituting this int

Vedclass · Accepted Answer

$\left( \frac{1}{\sqrt{2}}, 1 \right)$

Vedclass · Answer

(B) Given the inequality $\sin^{-1} x > \cos^{-1} x$ for $x \in (0, 1)$.We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$.Substituting this into the inequality,we get:$\sin^{-1} x > \frac{\pi}{2} - \sin^{-1} x$$2 \sin^{-1} x > \frac{\pi}{2}$$\sin^{-1} x > \frac{\pi}{4}$Since the sine function is strictly increasing on the interval $[0, 1]$,we apply the sine function to both sides:$x > \sin\left( \frac{\pi}{4} \right)$$x > \frac{1}{\sqrt{2}}$Given the domain $x \in (0, 1)$,the solution set is $x \in \left( \frac{1}{\sqrt{2}}, 1 \right)$.

Let $x \in (0, 1)$. The set of all $x$ such that $\sin^{-1} x > \cos^{-1} x$ is the interval

Similar Questions

The value of $\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is

The sum of possible values of $x$ for $\tan ^{-1}(x+1)+\cot ^{-1}\left(\frac{1}{x-1}\right)=\tan ^{-1}\left(\frac{8}{31}\right)$ is

If $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \frac{3\pi}{2}$,then the value of $x^{100} + y^{100} + z^{100} - \frac{9}{x^{101} + y^{101} + z^{101}}$ is equal to

$\tan ^{-1} 2 + \tan ^{-1} 3 = $

If $\alpha$ and $\beta$ are the least and the greatest values of $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$ for all $x \in [-1, 1]$ respectively,then $8(\alpha+\beta)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module