Equation {sec ^2}θ = frac{{4xy}}{{{{(x + y)}^2}}} is only possible when

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3.Trigonometrical Ratios, Functions and Identities

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

A

$x = y$

B

$x < y$

C

$x > y$

D

None of these

(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

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