The equation {(a + b)^2} = 4ab,{sin ^2}theta | vedclass

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Question

(b) We have ${(a + b)^2} = 4ab{\sin ^2}	heta $

$ \Rightarrow {\sin ^2}	heta = \frac{{{{(a + b)}^2}}}{{4ab}} \le 1 $

$\Rightarrow {(a + b)^2} -

Vedclass · Accepted Answer

$a = b$

Vedclass · Answer

(b) We have ${(a + b)^2} = 4ab{\sin ^2}	heta $

$ \Rightarrow {\sin ^2}	heta = \frac{{{{(a + b)}^2}}}{{4ab}} \le 1 $

$\Rightarrow {(a + b)^2} - 4ab \le 0$

$ \Rightarrow {(a - b)^2} \le 0 $

$\Rightarrow a = b.$

The equation ${(a + b)^2} = 4ab\,{\sin ^2}\theta $ is possible only when

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The equation ${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}}$ is only possible when