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3.Trigonometrical Ratios, Functions and Identities
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સમીકરણ ${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}}$ તો જ શક્ય છે જો . . . .
A
$x = y$
B
$x < y$
C
$x > y$
D
એકપણ નહીં.
(IIT-1966)
Solution
(a) Since ${\cos ^2}\theta \le 1$
${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}} \ge 1 $
$\Rightarrow 4xy \ge {(x + y)^2} $
$\Rightarrow {(x – y)^2} \le 0$
It is possible only when $x = y$, .$(x,\,y \in R)$
Standard 11
Mathematics
