operatorname{sech}^{-1}left(frac{3}{5}right)- | vedclass

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

Question

(C) We need to evaluate $\operatorname{sech}^{-1}\left(\frac{3}{5}ight)-	anh ^{-1}\left(\frac{3}{5}ight)$.First,we use the identity $\operatornam

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We need to evaluate $\operatorname{sech}^{-1}\left(\frac{3}{5}ight)-	anh ^{-1}\left(\frac{3}{5}ight)$.First,we use the identity $\operatorname{sech}^{-1}(x) = \cosh^{-1}\left(\frac{1}{x}ight)$.So,$\operatorname{sech}^{-1}\left(\frac{3}{5}ight) = \cosh^{-1}\left(\frac{5}{3}ight)$.Using the logarithmic form $\cosh^{-1}(x) = \log_e(x + \sqrt{x^2 - 1})$:$\cosh^{-1}\left(\frac{5}{3}ight) = \log_e\left(\frac{5}{3} + \sqrt{\frac{25}{9} - 1}ight) = \log_e\left(\frac{5}{3} + \sqrt{\frac{16}{9}}ight) = \log_e\left(\frac{5}{3} + \frac{4}{3}ight) = \log_e(3)$.Next,we use the identity $	anh^{-1}(x) = \frac{1}{2} \log_e\left(\frac{1+x}{1-x}ight)$.So,$	anh^{-1}\left(\frac{3}{5}ight) = \frac{1}{2} \log_e\left(\frac{1 + 3/5}{1 - 3/5}ight) = \frac{1}{2} \log_e\left(\frac{8/5}{2/5}ight) = \frac{1}{2} \log_e(4) = \frac{1}{2} \log_e(2^2) = \log_e(2)$.Finally,subtracting the two values:$\operatorname{sech}^{-1}\left(\frac{3}{5}ight) - 	anh^{-1}\left(\frac{3}{5}ight) = \log_e(3) - \log_e(2) = \log_e\left(\frac{3}{2}ight)$.

$\operatorname{sech}^{-1}\left(\frac{3}{5}\right)-\tanh ^{-1}\left(\frac{3}{5}\right)=$

Similar Questions

If $\sqrt{x + iy} = \pm(a + ib)$,then $\sqrt{-x - iy}$ is equal to

$\sqrt{3 + \sqrt{5}} - \sqrt{2 + \sqrt{3}} = $

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

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

$\sqrt{12-\sqrt{68+48 \sqrt{2}}}$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module