The value of x for which sin(cot^{-1} (1 + x) | vedclass

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

Question

(D) We are given the equation: $\sin(\cot^{-1}(1 + x)) = \cos(	an^{-1} x)$.First,simplify the right-hand side: Let $	an^{-1} x = 	heta$,then $x = \

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(D) We are given the equation: $\sin(\cot^{-1}(1 + x)) = \cos(	an^{-1} x)$.First,simplify the right-hand side: Let $	an^{-1} x = 	heta$,then $x = 	an 	heta$. Since $\cos 	heta = \frac{1}{\sec 	heta} = \frac{1}{\sqrt{1 + 	an^2 	heta}}$,we have $\cos(	an^{-1} x) = \frac{1}{\sqrt{1 + x^2}}$.Next,simplify the left-hand side: Let $\cot^{-1}(1 + x) = \phi$,then $1 + x = \cot \phi$. Since $\sin \phi = \frac{1}{\csc \phi} = \frac{1}{\sqrt{1 + \cot^2 \phi}}$,we have $\sin(\cot^{-1}(1 + x)) = \frac{1}{\sqrt{1 + (1 + x)^2}}$.Equating both sides: $\frac{1}{\sqrt{1 + (1 + x)^2}} = \frac{1}{\sqrt{1 + x^2}}$.Squaring both sides: $1 + (1 + x)^2 = 1 + x^2$.$(1 + x)^2 = x^2$.$1 + 2x + x^2 = x^2$.$1 + 2x = 0$.$x = -\frac{1}{2}$.

The value of $x$ for which $\sin(\cot^{-1} (1 + x)) = \cos(\tan^{-1} x)$ is

Similar Questions

If $4 \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$,then $x =$ . . . . . . .

If $\tan ^{-1}\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right]=\alpha$,then the value of $\sin 2 \alpha$ is

Find $\frac{dy}{dx}$ for $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right)$,where $-1 < x < 1$.

Let $(a, b) \subset (0, 2\pi)$ be the largest interval for which $\sin^{-1}(\sin \theta) - \cos^{-1}(\sin \theta) > 0$ holds for $\theta \in (0, 2\pi)$. If $\alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0$ and $\alpha - \beta = b - a$,then $\alpha$ is equal to:

$\sec ^2(\tan ^{-1} 2)+\operatorname{cosec}^2(\cot ^{-1} 3) = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module