sinh[log(2+sqrt{5})] + cosh[log(2+sqrt{3})] = | vedclass

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

Question

(A) We know that $\sinh(x) = \frac{e^x - e^{-x}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$.For the first term: $\sinh[\log(2+\sqrt{5})] = \frac{e^{\

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) We know that $\sinh(x) = \frac{e^x - e^{-x}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$.For the first term: $\sinh[\log(2+\sqrt{5})] = \frac{e^{\log(2+\sqrt{5})} - e^{-\log(2+\sqrt{5})}}{2} = \frac{(2+\sqrt{5}) - \frac{1}{2+\sqrt{5}}}{2}$.Since $\frac{1}{2+\sqrt{5}} = \sqrt{5}-2$,we have $\frac{(2+\sqrt{5}) - (\sqrt{5}-2)}{2} = \frac{4}{2} = 2$.For the second term: $\cosh[\log(2+\sqrt{3})] = \frac{e^{\log(2+\sqrt{3})} + e^{-\log(2+\sqrt{3})}}{2} = \frac{(2+\sqrt{3}) + \frac{1}{2+\sqrt{3}}}{2}$.Since $\frac{1}{2+\sqrt{3}} = 2-\sqrt{3}$,we have $\frac{(2+\sqrt{3}) + (2-\sqrt{3})}{2} = \frac{4}{2} = 2$.Adding these results: $2 + 2 = 4$.

$\sinh[\log(2+\sqrt{5})] + \cosh[\log(2+\sqrt{3})] = ?$

Similar Questions

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

The number of integers satisfying the inequality $\sqrt {{{\log }_3}(x) - 1} + \frac{{\frac{1}{2}{{\log }_3}({x^3})}}{{{{\log }_3}(\frac{1}{3})}} + 2 > 0$ is

The solution to the equation ${9^x} - {2^{x + 1/2}} = {2^{x + 3/2}} - {3^{2x - 1}}$ is:

The value of $\frac{\log_{3} 5 \times \log_{25} 27 \times \log_{49} 7}{\log_{81} 3}$ is

The roots of the equation $2^{x + 2} \cdot 27^{x/(x - 1)} = 9$ are given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module