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Question

$$\dfrac{\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...\infty }{1+\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+...\infty }$$ is equal to


A
e+1e1
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B
e1e+1
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C
e2+1e21
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D
e21e2+1
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Solution

The correct option is D $$\dfrac{e-1}{e+1}$$
$$\dfrac{\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...\infty }{1+\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+...\infty }$$

$$=\dfrac{2\left \{ \frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...\infty \right \}}{2\left \{ 1+\frac{1}{3!}+\frac{1}{5!}+...\infty  \right \}}=\dfrac{(e+e^{-1})-2}{(e-e^{-1})}$$

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

$$=\dfrac{(e-1)^2}{(e-1)(e+1)}$$

$$=\dfrac{e-1}{e+1}$$

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