sec h^{-1}left(frac{1}{2}right)-operatorname{ | vedclass

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

Question

(C) We use the logarithmic forms of inverse hyperbolic functions:$\sec h^{-1} x = \log _e\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cos

Vedclass · Accepted Answer

$\log _e\left(\frac{2+\sqrt{3}}{3}\right)$

Vedclass · Answer

(C) We use the logarithmic forms of inverse hyperbolic functions:$\sec h^{-1} x = \log _e\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cosec} h^{-1} x = \log _e\left(\frac{1+\sqrt{1+x^2}}{x}\right)$.For $\sec h^{-1}\left(\frac{1}{2}\right)$:$\sec h^{-1}\left(\frac{1}{2}\right) = \log _e\left(\frac{1+\sqrt{1-(1/2)^2}}{1/2}\right) = \log _e\left(\frac{1+\sqrt{3/4}}{1/2}\right) = \log _e\left(\frac{1+\sqrt{3}/2}{1/2}\right) = \log _e(2+\sqrt{3})$.For $\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$:$\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right) = \log _e\left(\frac{1+\sqrt{1+(3/4)^2}}{3/4}\right) = \log _e\left(\frac{1+\sqrt{25/16}}{3/4}\right) = \log _e\left(\frac{1+5/4}{3/4}\right) = \log _e\left(\frac{9/4}{3/4}\right) = \log _e(3)$.Subtracting the two values:$\log _e(2+\sqrt{3}) - \log _e(3) = \log _e\left(\frac{2+\sqrt{3}}{3}\right)$.

$\sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$ is equal to

Similar Questions

The square roots of the complex number $(-5-12i)$ are

$\sqrt{-8 - 6i} = $

If $z = \sum_{n=0}^{2026} i^n$, where $i = \sqrt{-1}$, then one of the values of $\sqrt{z}$ is...

$\sinh^{-1}\left(2^{3/2}\right)$ is equal to :

If the value of $\sqrt{-5-12 i}+\sqrt{7+24 i}$ is a negative real number $k$,then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module