operatorname{coth}^{-1} 3 + tanh^{-1} frac{1} | vedclass

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

Question

(B) We use the logarithmic definitions of inverse hyperbolic functions: $\operatorname{coth}^{-1} x = \frac{1}{2} \log_e \left(\frac{x+1}{x-1}\right)$

Vedclass · Accepted Answer

$\log_e 2\sqrt{3}$

Vedclass · Answer

(B) We use the logarithmic definitions of inverse hyperbolic functions: $\operatorname{coth}^{-1} x = \frac{1}{2} \log_e \left(\frac{x+1}{x-1}ight)$ for $|x| > 1$. $	anh^{-1} x = \frac{1}{2} \log_e \left(\frac{1+x}{1-x}ight)$ for $|x| $\operatorname{cosech}^{-1} x = \log_e \left(\frac{1 + \sqrt{1+x^2}}{x}ight)$ for $x > 0$ and $\operatorname{cosech}^{-1} x = \log_e \left(\frac{1 - \sqrt{1+x^2}}{x}ight)$ for $x Step $1$: Calculate each term. $\operatorname{coth}^{-1}(3) = \frac{1}{2} \log_e \left(\frac{3+1}{3-1}ight) = \frac{1}{2} \log_e(2) = \log_e \sqrt{2}$. $	anh^{-1}\left(\frac{1}{3}ight) = \frac{1}{2} \log_e \left(\frac{1+1/3}{1-1/3}ight) = \frac{1}{2} \log_e \left(\frac{4/3}{2/3}ight) = \frac{1}{2} \log_e(2) = \log_e \sqrt{2}$. $\operatorname{cosech}^{-1}(-\sqrt{3}) = \log_e \left(\frac{1 - \sqrt{1+(-\sqrt{3})^2}}{-\sqrt{3}}ight) = \log_e \left(\frac{1 - \sqrt{4}}{-\sqrt{3}}ight) = \log_e \left(\frac{1-2}{-\sqrt{3}}ight) = \log_e \left(\frac{-1}{-\sqrt{3}}ight) = \log_e \left(\frac{1}{\sqrt{3}}ight) = -\log_e \sqrt{3}$. Step $2$: Combine the terms. $\operatorname{coth}^{-1}(3) + 	anh^{-1}\left(\frac{1}{3}ight) - \operatorname{cosech}^{-1}(-\sqrt{3}) = \log_e \sqrt{2} + \log_e \sqrt{2} - (-\log_e \sqrt{3}) = \log_e \sqrt{2} + \log_e \sqrt{2} + \log_e \sqrt{3} = \log_e (\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3}) = \log_e (2\sqrt{3})$.

$\operatorname{coth}^{-1} 3 + \tanh^{-1} \frac{1}{3} - \operatorname{cosech}^{-1}(-\sqrt{3}) = $

Similar Questions

If $(x^2 \log _x 27) \cdot \log _9 x = x + 4$, then the value of $x$ is

$x = \log \left( \frac{1}{y} + \sqrt{1 + \frac{1}{y^2}} \right) \Rightarrow y$ is equal to

If $\log _{2}\left(9^{x-1}+7\right)-\log _{2}\left(3^{x-1}+1\right)=2$,then $x$ values are

The number of solutions to the equation $\log _2(x + 5) = 6 - x$ is

The interval of $x$ in which the inequality $5^{\frac{1}{4}(\log_5 x)^2} \geq 5x^{\frac{1}{5}(\log_5 x)}$ holds is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module