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 side: Let $	an^{-1} x = 	heta$,then $x = 	an

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 side: Let $	an^{-1} x = 	heta$,then $x = 	an 	heta$. Thus,$\cos(	an^{-1} x) = \cos 	heta = \frac{1}{\sec 	heta} = \frac{1}{\sqrt{1 + 	an^2 	heta}} = \frac{1}{\sqrt{1 + x^2}}$.Next,simplify the left side: Let $\cot^{-1} (1 + x) = \phi$,then $1 + x = \cot \phi$. Thus,$\sin(\cot^{-1} (1 + x)) = \sin \phi = \frac{1}{\csc \phi} = \frac{1}{\sqrt{1 + \cot^2 \phi}} = \frac{1}{\sqrt{1 + (1 + x)^2}}$.Equating both sides,we get: $\frac{1}{\sqrt{1 + (1 + x)^2}} = \frac{1}{\sqrt{1 + x^2}}$.Squaring both sides:$1 + (1 + x)^2 = 1 + x^2$$(1 + x)^2 = x^2$$1 + 2x + x^2 = x^2$$1 + 2x = 0$$2x = -1$$x = -\frac{1}{2}$.

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

Similar Questions

$\log _4 2 - \log _8 2 + \log _{16} 2 - \dots$ is equal to

The bond length of $HCl$ molecule is $1.275 \ \mathring{A}$ and its dipole moment is $1.03 \ D$. The ionic character of the molecule (in percent) (charge of the electron $= 4.8 \times 10^{-10} \ esu$) is

Which one of the following equations is correct for the reaction $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$?

The contribution in the total current flowing through a semiconductor due to electrons and holes are $3/4$ and $1/4$ respectively. If the drift velocity of electrons is $5/2$ times that of holes at this temperature,then the ratio of concentration of electrons and holes is

The study of the interaction of antigens and antibodies in the blood is known as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module