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{co

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 $\operatorname{cosec} h^{-1} x = \log _e\left(\frac{1+\sqrt{1+x^2}}{x}\right)$. Substituting $x = \frac{1}{2}$ in $\sec h^{-1} x$: $\sec h^{-1}\left(\frac{1}{2}\right) = \log _e\left(\frac{1+\sqrt{1-\frac{1}{4}}}{\frac{1}{2}}\right) = \log _e\left(\frac{1+\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \log _e(2+\sqrt{3})$. Substituting $x = \frac{3}{4}$ in $\operatorname{cosec} h^{-1} x$: $\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right) = \log _e\left(\frac{1+\sqrt{1+\frac{9}{16}}}{\frac{3}{4}}\right) = \log _e\left(\frac{1+\frac{5}{4}}{\frac{3}{4}}\right) = \log _e\left(\frac{\frac{9}{4}}{\frac{3}{4}}\right) = \log _e(3)$. Therefore,$\sec h^{-1}\left(\frac{1}{2}\right) - \operatorname{cosec} h^{-1}\left(\frac{3}{4}\right) = \log _e(2+\sqrt{3}) - \log _e(3) = \log _e\left(\frac{2+\sqrt{3}}{3}\right)$.

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

Similar Questions

The wavelength of the first line in the Balmer series in the hydrogen spectrum is $\lambda$. What is the wavelength of the second line?

If the length of the tangent from $(h, k)$ to the circle $x^2+y^2=16$ is twice the length of the tangent from the same point to the circle $x^2+y^2+2x+2y=0$,then

The maximum value of the function $f(x)=3x^3-18x^2+27x-40$ on the set $S=\{x \in R : x^2+30 \leq 11x\}$ is

$\lim _{x \rightarrow 0} \frac{\tan ^3 x-\sin ^3 x}{x^5}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module