Problem : If $\sin^2\theta = \frac{x^2+y^2+1}{2x}$ , $x$ must be .... Problem : If $\sin^2\theta = \frac{x^2+y^2+1}{2x}$ , $x$ must be 
(a) $1$ 
(b) $-2$ 
(c) $-3$ 
(d) $ 2$
My approach : 
Since $0 \leq \sin^2\theta \leq 1$ 
$\Rightarrow 0 \leq \frac{x^2+y^2+1}{2x} \leq 1 $
$\Rightarrow x^2 +y^2+1 -2x \leq 0$
$\Rightarrow (x-1)^2 +y^2 =0$ 
$\Rightarrow x =1 $ 
Therefore option is (a) 
Is it correct? please suggest thanks.

Yes the approach is right, except for one small thing, you should not just assume x to be positive while transferring variables across inequality signs. However, in this case, in the first inequality condition, if x is negative, the condition becomes:

$x^2 + y^2 + 1 <= 0$

which is wrong
x has to be positive, so doesn't really matter.
