A) \[\frac{e+1}{e-1}\]
B) \[\frac{e-1}{e+1}\]
C) \[\frac{{{e}^{2}}+1}{{{e}^{2}}-1}\]
D) \[\frac{{{e}^{2}}-1}{{{e}^{2}}+1}\]
Correct Answer: B
Solution :
\[\frac{\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...\infty }{1+\frac{1}{3!}+\frac{1}{5!}+...\infty }\] \[=\frac{2\left\{ \frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...\infty \right\}}{2\left\{ 1+\frac{1}{3!}+\frac{1}{5!}+...\infty \right\}}=\frac{(e+{{e}^{-1}})-2}{(e-{{e}^{-1}})}\] \[=\frac{e+\frac{1}{e}-2}{e-\frac{1}{e}}\] \[=\frac{{{e}^{2}}+1-2e}{{{e}^{2}}-1}=\frac{{{(e-1)}^{2}}}{(e-1)(e+1)}=\frac{e-1}{e+1}\]
You need to login to perform this action.
You will be redirected in
3 sec
