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

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