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3.Trigonometrical Ratios, Functions and Identities
medium
समीकरण ${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}}$ तभी सम्भव है जब
A
$x = y$
B
$x < y$
C
$x > y$
D
इनमें से कोई नहीं
(IIT-1966)
Solution
(a)चूँकि ${\cos ^2}\theta \le 1$
${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}} \ge 1 $
$\Rightarrow 4xy \ge {(x + y)^2} $
$\Rightarrow {(x – y)^2} \le 0$
$x = y$, $(x,\,y \in R)$
Standard 11
Mathematics
