Question Number 178355 by Spillover last updated on 15/Oct/22
$$\mathrm{Show}\:\mathrm{that}\: \\ $$ $$\mathrm{cosh}\:\mathrm{x}>\mathrm{sinh}\:\mathrm{x} \\ $$
Answered by Ar Brandon last updated on 16/Oct/22
$$\mathrm{cosh}{x}−\mathrm{sinh}{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\left({e}^{{x}} +{e}^{−{x}} \right)−\left({e}^{{x}} −{e}^{−{x}} \right)\right)={e}^{−{x}} >\mathrm{0} \\ $$ $$\Rightarrow\mathrm{cosh}{x}>\mathrm{sinh}{x} \\ $$
Commented bySpillover last updated on 16/Oct/22
$${thanks} \\ $$