Question

$se{c}^2\theta =\frac{4xy}{(x+y{)}^2}$ where $x\in R,y\in R$ is true if and only if:

• A. $x+y\ne 0$
• B. $x=y,x\ne 0$
• C. $x=y$
• D. $x\ne 0,y\ne 0$

Answer 1

Answer: (A), (C)

The correct options are
A $x+y\ne 0$
C $x=y$

For the above to be true
$4xy\ge (x+y{)}^2$ and $x+y\ne 0$

$(x+y{)}^2-4xy\le 0$
$(x-y{)}^2\le 0$

Now
$(x-y{)}^2<0$ this is not possible for real values of x and y.

Hence
$(x-y{)}^2=0$

$x-y=0$

$x=y$