MCQ
The equation ${(a + b)^2} = 4ab\,{\sin ^2}\theta $ is possible only when
- A$2a = b$
- ✓$a = b$
- C$a = 2b$
- DNone of these
$ \Rightarrow {\sin ^2}\theta = \frac{{{{(a + b)}^2}}}{{4ab}} \le 1 $
$\Rightarrow {(a + b)^2} - 4ab \le 0$
$ \Rightarrow {(a - b)^2} \le 0 $
$\Rightarrow a = b.$
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The sum $\sum_{n=4}^{\infty}\left(\frac{2 S_{n}}{n !}-\frac{1}{(n-2) !}\right)$ is equal to :