If x geq 1,then 2 tan^{-1} x + sin^{-1} (frac | vedclass

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

Question

(D) We are given the expression $f(x) = 2 	an^{-1} x + \sin^{-1} (\frac{2x}{1+x^2})$.We know the identity for $2 	an^{-1} x$ in terms of $\sin^{-1}$

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(D) We are given the expression $f(x) = 2 	an^{-1} x + \sin^{-1} (\frac{2x}{1+x^2})$.We know the identity for $2 	an^{-1} x$ in terms of $\sin^{-1}$ is:$2 	an^{-1} x = \begin{cases} \sin^{-1} (\frac{2x}{1+x^2}) & 	ext{if } |x| \leq 1 \ \pi - \sin^{-1} (\frac{2x}{1+x^2}) & 	ext{if } x > 1 \ -\pi - \sin^{-1} (\frac{2x}{1+x^2}) & 	ext{if } x For $x = 1$,$2 	an^{-1} (1) = 2(\frac{\pi}{4}) = \frac{\pi}{2}$ and $\sin^{-1} (\frac{2(1)}{1+1^2}) = \sin^{-1}(1) = \frac{\pi}{2}$. Thus,$f(1) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.For $x > 1$,we have $\sin^{-1} (\frac{2x}{1+x^2}) = \pi - 2 	an^{-1} x$.Substituting this into the expression:$f(x) = 2 	an^{-1} x + (\pi - 2 	an^{-1} x) = \pi$.Therefore,for all $x \geq 1$,the value of the expression is $\pi$.

If $x \geq 1$,then $2 \tan^{-1} x + \sin^{-1} (\frac{2x}{1+x^2})$ is equal to:

Similar Questions

Evaluate: $\sin ^{-1}\left[\sin \left(-600^{\circ}\right)\right]+\cot ^{-1}(-\sqrt{3})$

If $\tan ^{-1} x = \frac{\pi}{4} - \tan ^{-1} \left( \frac{1}{3} \right)$,then $x$ is

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin ^{-1} x + 3 \cos ^{-1} x = \frac{2 \pi}{5}$ is:

$\tan ^{-1} \sqrt{3} - \cot ^{-1}(-\sqrt{3}) = $ . . . . . . .

If $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$, then $x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module