Question

The solution of this division $(1+\frac{1}{2!}+\frac{1}{4!}+\dots .)÷(1+\frac{1}{3!}+\frac{1}{5!}+\dots ..)$ is equal to

• A. $\frac{e-1}{e+1}$
• B. $\frac{e+1}{e-1}$
• C. $\frac{e^2-1}{e^2+1}$
• D. $\frac{e^2+1}{e^2-1}$

Answer 1

Answer: (D)

$\frac{e^2+1}{e^2-1}$

The required expression $=\frac{(1+\frac{1}{2!}+\frac{1}{4!}---)}{(1+\frac{1}{3!}+\frac{1}{5!}---)}$
$=\frac{\frac{e+e^{-1}}{2}}{\frac{e-e^{-1}}{2}}$ .... $[e=1+\frac{1}{1!}+\frac{1}{2!}+....]$ and $[e^{-1}=1-\frac{1}{1!}+\frac{1}{2!}-....]$

$=\frac{e^2+1}{e^2-1}$