If the sum of all the solutions of tan ^{-1}l | vedclass

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

Question

(B) Given equation: $	an ^{-1}\left(\frac{2 x}{1-x^2}ight)+\cot ^{-1}\left(\frac{1-x^2}{2 x}ight)=\frac{\pi}{3}$.Since $\cot ^{-1}(y) = 	an ^{-1

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given equation: $	an ^{-1}\left(\frac{2 x}{1-x^2}ight)+\cot ^{-1}\left(\frac{1-x^2}{2 x}ight)=\frac{\pi}{3}$.Since $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$ for $y > 0$,we have $\cot ^{-1}(\frac{1-x^2}{2x}) = 	an ^{-1}(\frac{2x}{1-x^2})$ for $\frac{1-x^2}{2x} > 0$,i.e.,$x(1-x^2) > 0$.Case $I$: $x(1-x^2) > 0$. This holds for $x \in (-1, 0) \cup (0, 1)$.If $x \in (0, 1)$,then $\frac{2x}{1-x^2} > 0$,so $	an ^{-1}(\frac{2x}{1-x^2}) + 	an ^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{3} \Rightarrow 2 	an ^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{3} \Rightarrow 	an ^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{6}$.Then $\frac{2x}{1-x^2} = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \Rightarrow x^2 + 2\sqrt{3}x - 1 = 0$. Solving for $x > 0$,$x = \frac{-2\sqrt{3} + \sqrt{12+4}}{2} = 2 - \sqrt{3}$.Case $II$: $x(1-x^2) Then $\cot ^{-1}(\frac{1-x^2}{2x}) = \pi + 	an ^{-1}(\frac{2x}{1-x^2})$.The equation becomes $2 	an ^{-1}(\frac{2x}{1-x^2}) + \pi = \frac{\pi}{3} \Rightarrow 2 	an ^{-1}(\frac{2x}{1-x^2}) = -\frac{2\pi}{3} \Rightarrow 	an ^{-1}(\frac{2x}{1-x^2}) = -\frac{\pi}{3}$.Then $\frac{2x}{1-x^2} = 	an(-\frac{\pi}{3}) = -\sqrt{3} \Rightarrow \sqrt{3}x^2 - 2x - \sqrt{3} = 0$. Solving for $x The sum of solutions is $(2 - \sqrt{3}) + (-\frac{1}{\sqrt{3}}) = 2 - \frac{3+1}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}}$.Comparing with $\alpha - \frac{4}{\sqrt{3}}$,we get $\alpha = 2$.

If the sum of all the solutions of $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3}$ for $-1 < x < 1$ and $x \neq 0$ is $\alpha-\frac{4}{\sqrt{3}}$,then $\alpha$ is equal to $..........$.

Similar Questions

Let $f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}\right) + \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,then the value of $f(1) + f(2)$ is -

$\sec(\text{cosec}^{-1}x)$ is equal to

Find $\frac{dy}{dx}$ if $y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right)$,where $0 < x < \frac{1}{\sqrt{2}}$.

If $\frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1} \left(\frac{2^{p-1}}{1+2^{2p-1}}\right) = \tan^{-1} \alpha$, then $\tan \alpha$ is equal to . . . . . . .

If $2 \tan^{-1}(\cos x) = \tan^{-1}(2 \operatorname{cosec} x)$,then the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module