[JEE Advanced 1996] sec 2 theta 4xy x y 2 holds for real theta if and only
$\sec^2\theta=\dfrac{4xy}{(x+y)^2}$ holds (for real $\theta$) if and only if
(a) $x+y\ne0$
(b) $x=y,\ x\ne0$
(c) $x=y$
(d) $x\ne0,\ y\ne0$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
▲ 0
Correct answer: (b) $x=y,\ x\ne0$
$\sec^2\theta\ge1\Rightarrow\frac{4xy}{(x+y)^2}\ge1\Rightarrow(x-y)^2\le0\Rightarrow x=y$ (and $x\ne0$ so the fraction is defined and equals $1$).
JEE Advanced 1996 · Trigonometry — verified solution by the Vidaara Team.
Log in to post your own answer or join the discussion.
No comments yet — start the discussion.