cosh left(sinh ^{-1}(sqrt{8})+cosh ^{-1} 5rig | vedclass

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

Question

(B) Let $x = \sinh^{-1}(\sqrt{8})$ and $y = \cosh^{-1}(5)$.
Then $\sinh x = \sqrt{8}$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 8} = 3$.
Thus $

Vedclass · Accepted Answer

$15+8 \sqrt{3}$

Vedclass · Answer

(B) Let $x = \sinh^{-1}(\sqrt{8})$ and $y = \cosh^{-1}(5)$.
Then $\sinh x = \sqrt{8}$,so $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 8} = 3$.
Thus $e^x = \cosh x + \sinh x = 3 + \sqrt{8} = 3 + 2\sqrt{2}$.
For $y = \cosh^{-1}(5)$,$\cosh y = 5$ and $\sinh y = \sqrt{\cosh^2 y - 1} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$.
Thus $e^y = \cosh y + \sinh y = 5 + 2\sqrt{6}$.
We need to evaluate $\cosh(x + y) = \frac{e^{x+y} + e^{-(x+y)}}{2} = \frac{e^x e^y + e^{-x} e^{-y}}{2}$.
Note that $e^{-x} = \frac{1}{3 + 2\sqrt{2}} = 3 - 2\sqrt{2}$ and $e^{-y} = \frac{1}{5 + 2\sqrt{6}} = 5 - 2\sqrt{6}$.
$e^x e^y = (3 + 2\sqrt{2})(5 + 2\sqrt{6}) = 15 + 6\sqrt{6} + 10\sqrt{2} + 8\sqrt{3}$.
$e^{-x} e^{-y} = (3 - 2\sqrt{2})(5 - 2\sqrt{6}) = 15 - 6\sqrt{6} - 10\sqrt{2} + 8\sqrt{3}$.
$\cosh(x + y) = \frac{(15 + 6\sqrt{6} + 10\sqrt{2} + 8\sqrt{3}) + (15 - 6\sqrt{6} - 10\sqrt{2} + 8\sqrt{3})}{2} = \frac{30 + 16\sqrt{3}}{2} = 15 + 8\sqrt{3}$.

$\cosh \left(\sinh ^{-1}(\sqrt{8})+\cosh ^{-1} 5\right)=$

Similar Questions

If $\cot(\cos^{-1} x) = \sec(\tan^{-1} \frac{a}{\sqrt{b^2 - a^2}})$, then the value of $x$ is

If $y = \tan^{-1}\left(\frac{1}{x^2 + x + 1}\right) + \tan^{-1}\left(\frac{1}{x^2 + 3x + 3}\right) + \tan^{-1}\left(\frac{1}{x^2 + 5x + 7}\right) + \dots$ up to $n$ terms,then $\frac{dy}{dx}$ is equal to

If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$,then the value of $x$ is

If $0 < x < 1$ and $\tan ^{-1}(1-x), \tan ^{-1} x$ and $\tan ^{-1}(1+x)$ are in $A$.$P$.,then $x^3$ is equal to : . . . . . . .

$S = \tan^{-1}\left( \frac{1}{n^2 + n + 1} \right) + \tan^{-1}\left( \frac{1}{n^2 + 3n + 3} \right) + \dots + \tan^{-1}\left( \frac{1}{1 + (n + 19)(n + 20)} \right)$,then $\tan S$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module