# Sec 2 X = 4 X Y ( X + Y ) 2 is True If and Only If - Mathematics

MCQ
$\sec^2 x = \frac{4xy}{(x + y )^2}$ is true if and only if

• x + y ≠ 0

• x = y, x ≠ 0

• x = y

• x ≠0, y ≠ 0

#### Solution

x = y, x ≠ 0

We have:

$\sec^2 x = \frac{4xy}{(x + y )^2}$

$\Rightarrow \frac{4xy}{(x + y )^2} \geq 1 \left[ \because \sec^2 x \geq 1 \right]$

$\Rightarrow 4xy\geq(x + y )^2$
$\Rightarrow 4xy \geq x^2 + y^2 + 2xy$
$\Rightarrow 2xy \geq x^2 + y^2$
$\Rightarrow \left( x - y \right)^2 \leq 0$
$\Rightarrow \left( x - y \right) \leq 0$
$\Rightarrow x = y$
$\text{ For }x = 0, \sec^2 x \text{ will not be defined,}$
$\Rightarrow x \neq 0$
$\therefore x = y$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 5 Trigonometric Functions
Q 14 | Page 42