If 2 tan^{-1} x = 3 sin^{-1} x and x neq 0,th | vedclass

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

Question

(D) Given $2 	an^{-1} x = 3 \sin^{-1} x$. Using the identity $2 	an^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} ight)$ and $3 \sin^{-1} x = \sin^{-

Vedclass · Accepted Answer

$\sqrt{17}$

Vedclass · Answer

(D) Given $2 	an^{-1} x = 3 \sin^{-1} x$. Using the identity $2 	an^{-1} x = \sin^{-1} \left( \frac{2x}{1+x^2} ight)$ and $3 \sin^{-1} x = \sin^{-1} (3x - 4x^3)$,we have: $\sin^{-1} \left( \frac{2x}{1+x^2} ight) = \sin^{-1} (3x - 4x^3)$ $\Rightarrow \frac{2x}{1+x^2} = 3x - 4x^3$ Since $x 
eq 0$,we can divide by $x$: $\frac{2}{1+x^2} = 3 - 4x^2$ $2 = (3 - 4x^2)(1 + x^2)$ $2 = 3 + 3x^2 - 4x^2 - 4x^4$ $4x^4 + x^2 - 1 = 0$ Using the quadratic formula for $x^2$: $x^2 = \frac{-1 \pm \sqrt{1^2 - 4(4)(-1)}}{2(4)} = \frac{-1 \pm \sqrt{17}}{8}$ Since $x^2 > 0$,we take $x^2 = \frac{\sqrt{17} - 1}{8}$. Then $8x^2 = \sqrt{17} - 1$,which implies $8x^2 + 1 = \sqrt{17}$.

If $2 \tan^{-1} x = 3 \sin^{-1} x$ and $x \neq 0$,then $8x^2 + 1 =$

Similar Questions

If $\alpha, \beta$ are the solutions of the equation $\operatorname{Sin}^{-1} x - \operatorname{Cos}^{-1} x = \operatorname{Sin}^{-1}(3x - 2)$ and $\alpha > \beta$,then $3\alpha + 4\beta =$

The value of $\frac{d}{dx} \tan^{-1} \left[ \frac{3a^2x - x^3}{a(a^2 - 3x^2)} \right]$ at $x = 0$ is

The value of $\sec ^2(\tan ^{-1} 2)+\operatorname{cosec}^2(\cot ^{-1} 3)$ is

$\sin \left[ \cos^{-1} \left( \frac{3}{5} \right) + \tan^{-1} 2 \right] = $

The sum of the values of $x$ satisfying the equation $\sin ^{-1}\left(\frac{3 x}{5}\right)+\sin ^{-1}\left(\frac{4 x}{5}\right)=\sin ^{-1}(x)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module