If operatorname{Sinh}^{-1} x = operatorname{C | vedclass

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

Question

(A) Given $\operatorname{Sinh}^{-1} x = \log(1+\sqrt{2})$.Using the definition $\operatorname{Sinh}^{-1} x = \log(x + \sqrt{x^2+1})$, we have $x + \sq

Vedclass · Accepted Answer

$67 \frac{1}{2}^{\circ}$

Vedclass · Answer

(A) Given $\operatorname{Sinh}^{-1} x = \log(1+\sqrt{2})$.Using the definition $\operatorname{Sinh}^{-1} x = \log(x + \sqrt{x^2+1})$, we have $x + \sqrt{x^2+1} = 1+\sqrt{2}$.Comparing the terms, we get $x = 1$.Given $\operatorname{Cosh}^{-1} y = \log(1+\sqrt{2})$.Using the definition $\operatorname{Cosh}^{-1} y = \log(y + \sqrt{y^2-1})$, we have $y + \sqrt{y^2-1} = 1+\sqrt{2}$.This implies $y = \sqrt{2}$.Now, we need to find $\operatorname{Tan}^{-1}(x+y) = \operatorname{Tan}^{-1}(1+\sqrt{2})$.We know that $	an(67.5^{\circ}) = 	an(\frac{135^{\circ}}{2}) = \frac{1-\cos(135^{\circ})}{\sin(135^{\circ})} = \frac{1 - (-1/\sqrt{2})}{1/\sqrt{2}} = \sqrt{2}+1$.Therefore, $\operatorname{Tan}^{-1}(1+\sqrt{2}) = 67.5^{\circ} = 67 \frac{1}{2}^{\circ}$.

If $\operatorname{Sinh}^{-1} x = \operatorname{Cosh}^{-1} y = \log(1+\sqrt{2})$, then $\operatorname{Tan}^{-1}(x+y) = $

Similar Questions

$\sin 5^{\circ} + \sin 10^{\circ} + \sin 15^{\circ} + \dots + \sin 350^{\circ} + \sin 355^{\circ}$ is equal to:

If $\tan \beta = \frac{n \sin \alpha \cos \alpha}{1 - n \cos^2 \alpha}$,then $\tan (\alpha + \beta) \cdot \cot \alpha =$

$\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ = $

$\cos^2 76^{\circ} + \cos^2 16^{\circ} - \cos 76^{\circ} \cos 16^{\circ}$ is equal to

If $x \sin^3 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$ and $x \sin \alpha - y \cos \alpha = 0$,then $x^2 + y^2 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module