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3.Trigonometrical Ratios, Functions and Identities
medium
The equation ${(a + b)^2} = 4ab\,{\sin ^2}\theta $ is possible only when
A
$2a = b$
B
$a = b$
C
$a = 2b$
D
None of these
Solution
(b) We have ${(a + b)^2} = 4ab{\sin ^2}\theta $
$ \Rightarrow {\sin ^2}\theta = \frac{{{{(a + b)}^2}}}{{4ab}} \le 1 $
$\Rightarrow {(a + b)^2} – 4ab \le 0$
$ \Rightarrow {(a – b)^2} \le 0 $
$\Rightarrow a = b.$
Standard 11
Mathematics