Question

The number of points inside or on the circle $x^2+y^2=4$ satisfying ${\tan }^{4}x+{\cot }^{4}x+1=3{\sin }^{2}y$ is

• A. one
• B. two
• C. four
• D. none of these

Answer 1

The correct option is C four

${\tan }^{4}x+{\cot }^{4}x+1=3{\sin }^{2}y$
$({\tan }^{2}x+{\cot }^{2}x{)}^2-2{\tan }^{2}x.{\cot }^{2}x+1=3{\sin }^{2}y$
${({\tan }^{2}x+\frac{1}{{\tan }^{2}x})}^2-1=3{\sin }^{2}y$ (i)
Now using $A.M\ge G.M$
$\frac{{\tan }^{2}x+\frac{1}{{\tan }^{2}x}}{2}\ge {({\tan }^{2}x.\frac{1}{{\tan }^{2}x})}^{1/2}=1$
${\tan }^{2}x+\frac{1}{{\tan }^{2}x}\ge 2$
Hence minimum value of L.H.S of (i) is 3 Also $-1\le \sin y\le$, so maximum value of R.H.S is 3.
Thus (i) will true only if ${\tan }^{2}x=1,{\sin }^{2}y=1$
$x=\pm \frac{\pi }{4},y=\pm \frac{\pi }{2}$ since $(x^2+y^2\le 4)$
Hence number of points inside the given circle is $2\times 2=4$