The number of values of x satisfying the equa | vedclass

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

Question

(C) Let the given equation be $\operatorname{Tan}^{-1}\left(x+\frac{\sqrt{2}}{x}\right)+\operatorname{Tan}^{-1}\left(x-\frac{\sqrt{2}}{x}\right)=\oper

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let the given equation be $\operatorname{Tan}^{-1}\left(x+\frac{\sqrt{2}}{x}ight)+\operatorname{Tan}^{-1}\left(x-\frac{\sqrt{2}}{x}ight)=\operatorname{Tan}^{-1}(x)$.Applying the formula $\operatorname{Tan}^{-1}(A) + \operatorname{Tan}^{-1}(B) = \operatorname{Tan}^{-1}\left(\frac{A+B}{1-AB}ight)$, we get:$\operatorname{Tan}^{-1}\left(\frac{x+\frac{\sqrt{2}}{x} + x-\frac{\sqrt{2}}{x}}{1-(x+\frac{\sqrt{2}}{x})(x-\frac{\sqrt{2}}{x})}ight) = \operatorname{Tan}^{-1}(x)$.Simplifying the expression inside the $\operatorname{Tan}^{-1}$ function:$\frac{2x}{1-(x^2 - \frac{2}{x^2})} = x$.$\frac{2x}{1-x^2 + \frac{2}{x^2}} = x$.Assuming $x 
eq 0$, we can divide by $x$:$\frac{2}{1-x^2 + \frac{2}{x^2}} = 1$.$2 = 1 - x^2 + \frac{2}{x^2}$.$x^2 - \frac{2}{x^2} + 1 = 0$.Let $t = x^2$, then $t - \frac{2}{t} + 1 = 0 \implies t^2 + t - 2 = 0$.$(t+2)(t-1) = 0$.Since $t = x^2$, $t$ must be positive, so $t = 1$, which means $x^2 = 1$, so $x = \pm 1$.Checking $x = 1$: $\operatorname{Tan}^{-1}(1+\sqrt{2}) + \operatorname{Tan}^{-1}(1-\sqrt{2}) = \operatorname{Tan}^{-1}(1)$. This is true as $\operatorname{Tan}^{-1}(1+\sqrt{2}) + \operatorname{Tan}^{-1}(1-\sqrt{2}) = \frac{3\pi}{8} + \frac{\pi}{8} = \frac{\pi}{2}$ is not correct, but using $\operatorname{Tan}^{-1}(A) + \operatorname{Tan}^{-1}(B) = \operatorname{Tan}^{-1}(\frac{A+B}{1-AB})$ holds for $AB < 1$. Here $AB = x^2 - 2/x^2 = 1 - 2 = -1 < 1$. Thus, $x=1$ and $x=-1$ are solutions. There are $2$ values.

The number of values of $x$ satisfying the equation $\operatorname{Tan}^{-1}\left(x+\frac{\sqrt{2}}{x}\right)+\operatorname{Tan}^{-1}\left(x-\frac{\sqrt{2}}{x}\right)=\operatorname{Tan}^{-1}(x)$ is:

Similar Questions

Evaluate the sum: $\sum_{k=1}^{2026} \sin^{-1}(\cos \frac{k\pi}{4})$.

Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $[-1, 1]$ such that $\cos ^{-1} x - \sin ^{-1} y = \alpha$,where $-\frac{\pi}{2} \leq \alpha \leq \pi$. Then,the minimum value of $x^2 + y^2 + 2xy \sin \alpha$ is

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

The number of real solutions of the equation $\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2+x+1} = \pi/2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module